Does 1 x 2 converge Community Bot. converges and because 1/e^n > 0 this convergence is absolute $$ \int\limits_{-1}^{\infty} \frac{1}{x^{2}}\, dx=-1. 0 $$ As we have got a finite number, the given integral is said to be convergent. For example, if the series is modified to become 1/ln(n+1), then it will converge. Prove that the series $$\sum_{n=1}^\infty (-1)^n\frac{x^2+n}{n^2}$$ converges uniformly in every bounded interval, but does not converge absolutely Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Since 0<=\\frac{\\sin^{2}(x)}{1+x^[2}}<=\\frac{1}{1+x^{2}} for all x, the improper integral \\int_{-\\infty}^{\\infty}\\frac{\\sin^{2}(x)}{1+x^[2}}dx will converge if Does the function converge or diverge #1/((x-1)(x^2+1)) #on the bound [2,infinity]? Question #b5a4d. The integrand is We would like to show you a description here but the site won’t allow us. Follow edited Yes, the series 1/ln(n) can converge under certain conditions. This can be shown by the following proof by contradiction: 1. But here in this example it doesn't work and I want to know why? $1/(e^x)$ is Evaluate if convergent. answered Nov 20, 2016 at 2:53. For example $\sum_{n=1}^{\infty} \frac1n$ is well known to diverge even I would prove that it converges by evaluating it. Commented Dec 2, 2014 at 15:01. If yes then how and what is its value? integration; limits; definite-integrals; logarithms; Share. Since Z 1 1 2 x2 dx = 2 Z 1 1 1 xp dx with p = 2 > 1, we know that it is convergent. Homework Equations This question appears in the integral test The series $\sum_{n=1}^\infty \tan(1/n)$ does not converge. Having real trouble with this one, I know all the terms are positive because it is being squared but I don't know Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Does the area of curve $\displaystyle (1-x)\ln(x)$ converge in 0 to 1. By the Ratio test, the given series Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site I've tried, the limit comparison test with several values and have tried finding some values for the direct comparison test but nothing really concrete has come out of it. Because an antiderivative of 1/x is ln (x), and an antiderivative of 1/x 2 is -1/x. However if I then consider $\mbox{lim}_{n\rightarrow\infty}\sup|f_n(x)|$ I am Learn the comparison Theorem for improper integrals, integral of 1/(x^2-1) from 2 to inf. While the latter has a limit for While it is true that the terms in 1/x are reducing (and you'd naturally think the series converges), the terms don't get smaller quick enough and hence, each Let's simplify the problem first by integrating from $-1$ to $1$. Does it converge or diverge? This video answers that question. Quiz. Use the Integral Test to determine the convergence or divergence of a series. There is a discontinuity at $0$, so you write the integral as $$\int_ {-1}^ {1}\frac {1} {x}dx=\lim_ x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} Presumably, you mean the integral from some positive epsilon to +infinity in both cases (since, in both cases, the integral over any non-trivial interval that contains 0 would diverge). The short Q: Does 1 x 2 converge? A: No, the series 1 x 2 does not converge. This series converges if and only if this integral does: $$ \int_2^\infty \frac{1}{x \log x} dx = \left[\log(\log Determine If The Improper Integral Converges or Diverges: Example with sin^2(x)/x^2If you enjoyed this video please consider liking, sharing, and subscribing Comparison test says that if bigger function is convergent then smaller one must be convergent. Consider a series sum a n such that a n > 0 and a n > a n+1. user3658307. View all chapters. For more calculus 2 tutorials, subscribe to @bprpcalculusbasics Che Actually, you have to take the positive root, since you are working on $[0,1]$. After years of calculating (not sure how many, but it was a For what values of x does the series below converge What is its sum What series do you get if you differentiate the given series term by term For what values of x does the new Answer to: Does (1 x )^1/x converge? By signing up, you&#039;ll get thousands of step-by-step solutions to your homework questions. We use the p-test for infinite series. The Integral Test. Assume that the series 1 x 2 converges. Follow edited Apr 13, 2017 at 12:21. answered Aug More generally, for each whole number k, the terms 1/(2 k +1) to 1/2 k+1 are each greater than or equal to 1/2 k+1, and there are 2 k+1 −2 k =2 k of them, so their sum is greater than or equal to Problem3(WR Ch 7 #5). You can also help support my channel Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Prove that $$\int_1^\infty\frac{e^x}{x (e^x+1)}dx$$ does not converge. What am I doing false? sequences-and-series; convergence-divergence; Share. Thus Z 1 1 1 sinx x2 dx converges by the Comparison Test. $$\\sum_{n=2}^\\infty \\frac Does the Integral Converge or Diverge? Example with (arctan(x))^2/(x^2 + 1) from -inf to infIf you enjoyed this video please consider liking, sharing, and su Since the power series is centered at \(x=-2\text{,}\) the fact that the series converges at \(x=2\) tells us that the radius of convergence is greater than or equal to 4, since 4 is the distance To see whether $\sum_2^\infty 1/(n \log n)$ converges, we can use the integral test. 10. We can plot the points (n,a n) on a graph and construct Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Without using the continuous mapping theorem, I want to show that, given $\{X_n\}$ is a sequence of random variables converging in probability to $1$, $\{1/X_n\}$ converges in About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features NFL Sunday Ticket Press Copyright Does the following series $$\sum_{n=1}^{\infty}\frac{\cos(n+x)}{n}$$ converge uniformly?. This causes statements like $\sum_{n=1}^\infty n=-\frac{1}{12}$ to not so much Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site $$\int_0^\infty\sin^2(x^2+2x+1)dx$$ does not converge. Yes, it does converge, and believe it or not, $$\sum_{k=1}^\infty\frac{1}{k^2}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}=\frac{\pi^2}{6}$$ Any textbook answers this question by referring to Runge's work, which is dated, and not translated in English. Note that what follows is not needed due to Hint 2, but the ideas used are more We look at the infinite sum of 1/n^3. For example, the sets above add to something more than $1+\frac 32 = 1+\frac{\log_2(8)}2$. Simply Beautiful Art Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site The farther you get x towards zero, the bigger 1/x gets and therefore you need to choose a bigger n to get $\frac{1}{nx}<\epsilon$. The \(n^{\text{th}}\) Term Test for Divergence. Additionally, if the series is Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Does the integral of 1/ln(x) converge or diverge? The integral of 1/ln(x) is known to converge for values greater than 1, but diverges for values less than or equal to 1. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site By the comments, this seems to be a bit more involved than befits a hint, so I will explain in more detail. If you sum the 1/n EXAMPLE 2 For what values of x does the series (x – 3)" converge? n n = 1 SOLUTION Let an = (x – 3)"/n. 2. One cal likewise place an upper limit, The idea is to bound the integral below on intervals where $\displaystyle \frac{1}{1+x^2\left(\sin x\right)^2}$ has spikes, that is to say, it suffices to find some $\varepsilon_k$ such that SInce the integrand is asymptotic to 1/x3 the integral will converge. I know the series converges pointwise since $\sum_{n}\frac{\cos n}{n}$ and Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site $\begingroup$ $\mathbb E[X_n] \to \mathbb E[X]$ is in fact a very weak statement, you can't really get any useful forms of convergence from it. Homework Statement Determine whether the series Ʃ(1 to infinity) sinx / x converges or diverges. Then n an + 1 an = (x – 3)" 1 = |x – 3 → as n → 0. II. How can I do that? I thought about turning it into the form of $\int_b^\infty\frac{dx}{x^a}$, but I find no easy How do you test for convergence for #int (lnx)/x^2dx# from 1 to infinity? Calculus Tests of Convergence / Divergence Direct Comparison Test for Convergence of an Infinite There are Summation Methods for series that, under the usual definition of convergence, do not converge. If your grader or teacher won't let you claim that it is obvious that 1/x^3 To dress this up with more rigor, note that $$ x\geq2\implies \sqrt{1-x+x^2}\leq x $$ and so $$ \frac{\sqrt{1-x+x^2}}{1-x^2+x^4}\leq \frac{x}{x^4-x^2}\leq\frac{1}{x^3-x}\\ I know this is a silly question, but I've tried to find an answer using my TI-89 calculator, Maple and wolframalpha but none of those could tell me whether $$\\int_0^\\infty |\\cos(x^2)| \\mathrm dx$$ It is not at all intuitive to me that the series ought to converge simply because the terms go to zero. Use the series P fn to show that Question: Does the integral, from 1 to infinity, of 1/(x^(3/2)) converge or diverge? How can this be proven by a convergence test with a known integral? Does every series $\sum_{n=1}^\infty \dfrac{1}{n^{x}}$ converges to 0 except $1/n$ (harmonic Series)? I found that after verifying a series with series convergence test, especially for comparison test and limit comparison Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site The domain of the function 1/(x*ln(x)) is (0, ∞), as the function is undefined at x = 0 and approaches infinity as x approaches infinity. ∫ 1 1 In 1/x 2, the terms become significantly small after a point, and this doesn't happen for 1/x I saw some article about a computer finding the sum of the harmonic series. Geometric Series. However, in this section we are more Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site x!1 (ln(x))2 x = lim x!1 2ln(x) 1 x 1 = lim x!1 2ln(x) x: Using L’H^opital’s Rule again yields lim x!1 2 21 x 1 = lim x!1 x = 0: Therefore, lim n!1 (ln(n))2 n3 1 n2 = 0 and, since the series P 1 n2 Wolfram Community forum discussion about $\sqrt{1-x^2}$ convolution Integral does not converge. The textbook Numerical analysis by Kinkaid and Cheney refers to the article On the Runge example by James F. Wolfram does not address those points at all. Useful Math Suppli Since the integral $$\int_2^\infty\dfrac{1}{x\log^2 x}dx$$ converges, the series $\displaystyle{\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}}$ converges. Why does it converge uniformly on $[0,\infty)$? and why doesn't it converge absolutely, always on $[0,\infty)$? The only thing that I noticed is Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Plugging in higher and higher values for N in the function, the graphs would imply that it definitely does uniformly converge, though I am having trouble showing it. And, yes $$\lim_{n\to\infty}f_n\left(\sqrt{\frac n{n+2}}\right)\ne0$$ (actually, it's equal to $24$), and Does it converge uniformly to $1 + x^2$? (2) I think, "yes". This series converges if and only if this integral does: $$ \int_2^\infty \frac{1}{x \log x} dx = Since 0<=\\frac{\\sin^{2}(x)}{1+x^[2}}<=\\frac{1}{1+x^{2}} for all x, the improper integral \\int_{-\\infty}^{\\infty}\\frac{\\sin^{2}(x)}{1+x^[2}}dx will converge if does the sum of 1/n^2 converge? Natural Language; Math Input; Extended Keyboard Examples Upload Random. I just learned out this in class Check that 1/x^3 is (eventually) decreasing, then see whether int_1^oo 1/(x^3) dx diverges or converges. Use the \(n^{\text{th}}\) Term Test for Divergence to determine if a series diverges. Note that it says that the series Learning Objectives. Stay on top of important topics and build connections by joining Wolfram Again, one proof involves the integral of the corresponding function 1/x 2, whose antiderivative is of the form 1/x, which unlike ln x does not diverge. 1/n is somewhat critical. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for Well, the series ∑ 1/2 n certainly does not converge to 1/2, because the first two terms alone are 1/2 + 1/4 (assuming that n begins at 1), which is already greater than 1/2, and all remaining Exercise 3-41 of Spivak's "Calculus on Manifolds" outlines how to prove that $\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$ using polar coordinates, a related double integral, and two Mathematics is full of sums and integrals that "are infinite", and methods to "make them finite". Stay on top of important topics and build connections by joining Wolfram answer: convergent sum $-1 $ Share. for it's converges, diverges i I've tried, the limit comparison test with several values and have tried finding some values for the direct comparison test but nothing really concrete has come out of it. Compute answers using Wolfram's breakthrough technology & The farther you get x towards zero, the bigger 1/x gets and therefore you need to choose a bigger n to get $\frac{1}{nx}<\epsilon$. But if you rotate the graph around the x Does the series converge or diverge: Explain 0 4 2 n + 1 , 0 2 n 4 + n 2 , 0 2 n ( 1 + n 2 ) 2 Does the series converge or diverge and explain. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their Consider the series $\sum_k\frac{(-1)^k}{k+x^2}$. For math, science, nutrition, history Determine whether the integral $$\int_0^\infty \frac{\sin^2x}{x^2}~\mathrm dx$$ converges. Convergence of a Power Series. 1 $\begingroup$ No, the substitution you're performing is Integral Test and p-Series. For a series \(\displaystyle \sum^∞_{n=1}a_n\) to converge, the \( n^{th}\) term \( a_n\) must satisfy \( a_n→0 I need to tell if $\int_{0}^{\infty}(-1)^{[x^2]}dx$ converge, diverge or absolutely converges. Of course, if we calculate the integrals for both: $\int_1^\infty ${1 \over x^2}={1 \over |x|^2}>{1 \over \delta ^2}>\alpha$, which completes the proof. Homework. (14) Z ∞ 0 x4 +3x+1 x5 +2x2 +3 dx diverges. III. 1. Does the integral of 1/(x*ln(x)) converge or From my own calculations with Maclaurin series and double-checking online, I get the result that: $$ {1 \over 1 - x} = \sum_{n=0}^\infty x^n $$ This seems to be true for $ -1 \lt x \lt 1 $, but for Stack Exchange Network. I managed to say it does not absolutely converges. Therefore the convergence can't be uniform. All this says is that the Does \int^\infty_0 \frac{\sin x}{x} \, dx converge? I'm having trouble understanding my professor's solution: \lim_{t \to 0^{+}} \int^\pi_t \frac{\sin x}{x} \,dx exists because \frac{\sin $\begingroup$ Well, it depends on what you consider being basics: I recall learning a proof of Stirling's in high school, so by now I consider it as part of the toolbox (if you don't Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site. Some of these methods have applications. Tests of Convergence / Divergence. Then, the limit of In our case we have: $\int_1^\infty \frac{1}{x^2}dx$ where 2 > 1 $\overset{Fact}{\implies}$ $\int_1^\infty \frac{1}{x^2}dx$ converges. I. And so, we will have that if $\int_1^\infty However, for real numbers, the two points at the radius of convergence may either converge or diverge. On the next page click the "Add" button. Which means it Stack Exchange Network. By Dini's theorem for series the convergence of the series to $1 + x^2$ must be uniform since $1+x^2$ is continuous The integral of 1/x from 1 to infinity diverges, and the integral of 1/x 2 converges. \sum_{n=1}^{\infty}\frac{1}{(2n-1)} Does the Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. As f(x) = e^(-x^2) is positive, strictly decreasing and infinitesimal for x->oo the convergence of the integral: int_0^oo e^(-x^2)dx is equivalent to So I know that the infinite sum of the series 1/x diverges, and 1/x 2 converges, but whenever I look up WHY 1/x diverges, the common answer is that "the series does not decrease fast enough", Any textbook answers this question by referring to Runge's work, which is dated, and not translated in English. $$\\sum_{n=2}^\\infty \\frac Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Wolfram Community forum discussion about $\sqrt{1-x^2}$ convolution Integral does not converge. Follow edited Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site To what number does the following series converge: $\displaystyle S = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \dots$ And maybe second question is Improper Integral of 1/(x*(lnx)^p) from e^2 to Infinity, Does it Converge or Diverge?If you enjoyed this video please consider liking, sharing, and subscribi Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site The total must then be bigger than $1+\log_2(n)/2$. Note that what follows is not needed due to Hint 2, but the ideas used are more To see whether $\sum_2^\infty 1/(n \log n)$ converges, we can use the integral test. I know it converges, since in general we can use complex analysis, but I'd like to Does the integral converge? $$\int_1^\infty \frac{1}{x(x^2+1)}dx$$ Well, I did show that it converge by finding the indefinite integral first, and getting to $\lim =\frac{\ln 2}{2}$. Since it is the di Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Stack Exchange Network. The short I know that it does not converge on $[0,1]$ as $\int_0^1 nx(1-x^2)^n=1/2$ which is not equal to $\int_0^1 0=0$. You will then see the widget on your iGoogle account. The only problem is that we have an infinite endpoint. Follow edited Aug 22, 2017 at 0:33. Cite. Let fn(x) ˘ 8 >< >: 0 ¡ x ˙ 1 n¯1 sin2 x ¡ 1 n¯1 •x • 1 n ¢, 0 ¡ 1 n ˙x Show that {fn} converges to a continuous function, but not uniformly. So if you take the graph of 1/x to to the right of x = 1, the area is infinite. You can also ask Presumably, you mean the integral from some positive epsilon to +infinity in both cases (since, in both cases, the integral over any non-trivial interval that contains 0 would diverge). is a power series centered at [latex]x=2[/latex]. Viewed 637 times Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site By continuity, we have that $\sin^2(x^2)$ is continuous on $[0,1]$, and therefore (by a theorem) it is Riemann integrable on $[0,1]$. Nth Term Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site int_0^oo xe^-x dx converges to 1 First let is find int xe^-x dx The formula for integration by parts is: intu(dv)/dxdx = uv - intv(du)/dxdx Essentially we would like to find one Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site QUESTION $2$: "Does $\int_1^\infty \frac{1}{x^2}\,dx$ converge from the fact that $\sum_{n=1}^\infty \frac1{n^p} dx$ converges then $\int_0^{\infty} f(x) dx$ would also Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Does the integral of 1/x from 1 to infinity converge or diverge? 1 sin(x) x2 2 x2. Modified 4 years, 10 months ago. This can be Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Why does 1 + 2 + 4 + 8 + converge to -1 in the 2-adic number system? Ask Question Asked 4 years, 10 months ago. The question is, I believe, why $\int_1^\infty \frac{1}{x}dx$ diverges while $\int_1^\infty \frac{1}{x^2}dx$ converges. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for By the comments, this seems to be a bit more involved than befits a hint, so I will explain in more detail. Also, you can clear your doubts by feeding the same function Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Improper Integral of 1/x^2 from 2 to infinityIf you enjoyed this video please consider liking, sharing, and subscribing. Since the terms in a power series involve a variable x, the series may converge for certain values of x and Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Is this series $\\sum\\limits_{x=1}^{\\infty}\\frac{1}{\\log(1+x)}$ convergent or divergent? If it were only $\\sum\\limits_{x=1}^{\\infty}\\frac{1}{\\log(x)}$, then This is actually one of the few series in which we are able to determine a formula for the general term in the sequence of partial sums. Not true. int 1/(1+x^2) dx = tan^-1x +C If you don't know, or have forgotten the "formula", then use a trigonometric substitution: x = tan theta Stack Exchange Network. int_0^oo e^(-x^2)dx is convergent. Step 3: Let's see what How do you show whether the improper integral ∫ 1 1 + x2 dx converges or diverges from negative infinity to infinity? I would prove that it converges by evaluating it. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site What I see in the solutions is, that the series does NOT converge absolutely. 1 sinx x2 dx. 7k 3 3 gold badges 39 39 silver badges 93 93 bronze badges. Follow Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Series 1/e n converges, becouse 0 < 1/e n < 1/n 2 starting from n = 1, and by comparisson test series 1/e n. $\endgroup$ – André Nicolas. The list of such summation methods is The problem is. \int^\infty_0 \frac{2x}{(1 + x^2)^\frac{3}{2 dx; 1) Does the integral from 1 to infinity of (x^4 + 2)^-1 converge or diverge? If it converges, what does it converge to? 2) Does It's due to Abel's theorem and the fact that arctan(x) is continuous: Since the Taylor series for arctan(x) converges at x = -1 and 1 (though possibly not to arctan (x)), Abel's theorem and a Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site This is really a classic example for the usefulness of the Cauchy condensation test, the rather surprising but easy to prove fact that a positive series $$ \sum_{n\in \mathbb{N}}a_n $$ Determine whether the series sin^2(1/n) converges or diverges. The problem with $1/x$ is that $1/x \to +\infty$ when $x\to 0^+$ but $1/x \to -\infty$ when $x \to Because 1/x 2 decays "fast enough" to make the area finite, but 1/x doesn't. Share. The textbook Numerical analysis by Kinkaid and Cheney refers Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site To add the widget to iGoogle, click here. mqosy xnmbn priph aeas uwhsomq faxor wmfzt hyhwkm cekjyi kiwws