$\sum_{n=1}^{+\infty}\left[n \log\left ( \frac{2n+1}{2n-1} \right )-1\right]$ ,converges or not  My attempt  $a_n=n \log\left ( \frac{2n+1}{2n-1} \right )-1 \\ \\ \log \left ( \frac{2n+1}{2n-1} \right )=\log \left ( 1+\frac{2}{2n-1} \right ) \sim \frac{2}{2n-1} \ (n\rightarrow +\infty) \\ \\ a_n=n\log \left ( 1+\frac{2}{2n-1} \right )-1 \sim n \frac{2}{2n-1}-1=\frac{2n}{2n-1}-1=\frac{1}{2n-1}  \\ \\ $  $\sum_{n=1}^{+\infty}\frac{1}{2n-1} $  diverges  ,then $\sum_{n=1}^{+\infty}a_n $ diverges   but according to Wolfram  $\sum_{n=1}^{+\infty}a_n=\frac{1}{2}-\frac{\log 2}{2}$  where is the mistake in my solution please?  thanks
73242
I am investigating the convergence of  $$\sum _{n=1}^{\infty }\left\{ 1-n\log \frac {2n+1} {2n-1}\right\} $$ Now as per Cauchy's test for absolute convergence.  If $\overline {\lim _\limits{n\rightarrow \infty }}\left| u_{n}\right|^{{1}/{n}} &lt; 1,\sum _\limits{n=1}^{\infty }u_{n}$ converges absolutely  Obviously, if $\overline {\lim \limits_{n\rightarrow \infty }}\left| u_{n}\right|^{{1}/{n}} &gt; 1,\sum _\limits{n=1}^{\infty }u_{n}$ does not converge.  I observed $$\overline {\lim _{n\rightarrow \infty }}\left| \log \left( \dfrac {2n+1} {2n-1}\right) ^{-n}\right|  = \overline {\lim _{n\rightarrow \infty }}\left| \log \left( 1-\dfrac {1} {n-{1}/{2}}\right) ^{-n}\right|$$  Could I take the negative power of $n$ outside the absolute brackets here? I guess even if I could establish $\log$ part converges that would only show that the overall series diverges right. Is that the correct result ? Any alternative lines of attacking this problem would be much appreciated.
21884
I am attempting to prove if the below expression converges or diverges:  $$\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}$$  I decided to try using the . So then, I set $f(n) = \frac{1}{n (\ln n)^2}$ and $g(n) = \frac{1}{n}$ and did the following:  $$ \lim_{n \to \infty} \frac{f'(n)}{g'(n)} =  \lim_{n \to \infty} \frac{-\frac{\ln(n) + 2}{n^2 (\ln n)^3}}{-\frac{1}{n^2}} =  \lim_{n \to \infty} \frac{\ln(n) + 2}{n^2 (\ln n)^3} \cdot n^2 =  \lim_{n \to \infty} \frac{1}{(\ln n)^2} + \frac{2}{(\ln n)^3} = 0 $$  I know that $\sum \frac{1}{n}$ diverges due to properties of harmonic series, and so concluded that my first expression $\frac{1}{n (\ln n)^2}$ also must diverge.  However, according to , and as suggested by , the expression actually converges.  What did I do wrong?
68289
Question is to check if : $\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}&gt;\frac{3}{5}$   the problem is that i am sure that this series$\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}$ is convergent (if it is not then i would be very happy as i will then be done.)  I would be thankful if someone can help me out with solving this.
119969
Suppose $\sum_{n=2}^{\infty} a_n$ is a convergent series. Is the following series convergent as well? $$\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)$$ The part with $n^{a_n}-1 $ looks quite similar to $\sqrt[n]{n}-1$ which behaves like $\ln{n}$, however, I'm not able to tie this fact to the behavior of the whole expression. Any suggestions?
23735
Show that $\frac{1}{s-1}+\sum_{n=1}^{\infty}\left[\frac{1}{n^s}+\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} - \frac{1}{n^{s-1}}\right)\right]$ converges when $0&lt;s&lt;1$.  Note that this series is equivalent to $\sum_{n=1}^{\infty}\frac{1}{n^s}$. It might seem counterintuitive that the series converges if $0&lt;s&lt;1$.
91006
$$\sum_{n=2}^\infty \frac{1}{(\ln\, n)^2}$$  The series converge?  Please verify my solution below
80108
I'd appreciate any help, I know it has something to do with the geometric series but I still can't figure out how. I thought about integration but couldn't find a way to do it. $$\sum_{n=1}^\infty \frac{1}{2^nn(3n-1)}$$  Thanks.
34609
how can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $? It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...  Let consider sum no to infinity, but to n  $$ \sum_{k=1}^{n} (k\cdot \ln \frac{2k+1}{2k-1} - 1) =$$ $$ ln \frac{3}{1}\cdot \left(\frac{5}{3}\right)^2 \cdot...\cdot \left(\frac{2n+1}{2n-1}\right)^n - n = ln \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1} - n $$ but $$ n = ln e^n $$ so it will be $$ln\frac{1}{e^n} \cdot \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1}$$  So the limit of it is $-\infty$ Have I done this well or I missed sth?
30147
Consider convergence of series:  $$\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$$  My tried:  We have  $$\sum_{n=1}^{\infty }\sin(\pi (2+\sqrt{3})^{n})=\sum_{n=1}^{\infty}\sin\left(\pi[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}]-\pi(2-\sqrt{3})^{n}\right)\, (*)$$ Because $$(2+\sqrt{3})^{n}=\sum_{k=0}^{n}C_{n}^{k}2^{n-k}3^{\frac{k}{2}}$$                 $$(2-\sqrt{3})^{n}=\sum_{k=0}^{n}(-1)^{k}C_{n}^{k}2^{n-k}3^{\frac{k}{2}}$$ Hence $$(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}=\left\{\begin{matrix} 0&amp;,k=2l+1 \\ m\in N&amp;,k=2l \end{matrix}\right.$$        $$\Rightarrow (1)=\sum_{n=1}^{\infty}\sin\left(m\pi-\pi(2-\sqrt{3})^{n}\right)=\sum_{n=1}^{\infty}(-1)^{m+1}\sin\frac{\pi}{(2+\sqrt{3})^{n}}&lt;\sum_{n=1}^{\infty}\sin\frac{\pi}{(2+\sqrt{3})^n}$$  $\sum_{n=1}^{\infty}\sin\frac{\pi}{(2+\sqrt{3})^n}$ converge  Hence series is converge.  True or False?
19661
I tried to use the absolute convergence test. $$\int_1^2\left|\frac{\sin{x}}{\log{x}}\right| \, dx&lt;\int_1^2\left|\frac{1}{\log{x}}\right| \, dx$$ but I'm not sure whether $\frac{1}{\log{x}}$ converges or not.  Thanks in advance!
36198
I noticed that a lot of the time, people ask whether the following sum converges:  $$\sum_{n=1}^\infty\frac{\sin(n)}n$$  Though I've never stopped to ask what it equaled.  According to this , the sum is given as  $$\sum_{n=1}^\infty\frac{\sin(n)}n=\frac{\pi-1}2$$  The solution involves realizing $\sin(n)=\Im e^{in}$ and the Taylor expansion for the natural logarithm.  While thats great and all, how can I prove this using only real numbers?
88762
I have a series here, and I'm supposed to determine whether it converges or diverges. I've tried the different tests, but I can't quite get the answer.     $$\sum_{n=1}^\infty\ln\left(1+\frac1{n^2}\right)$$
2553
Is this convergent or divergent? if convergent, find the sum. If divergent explain why.  $$\sum _{n=1}^{\infty }\:\frac{2n^2}{5n^2+2n+1}$$  I want to use the divergent test which is $\mathrm{If\:}\lim _{n\to \infty }a_n\ne 0\mathrm{\:then\:}\sum a_n\mathrm{\:diverges}$   $\frac{2n^2}{5n^2+2n+1}$  I would assume to take out the largest $n$ in both the top and bottom   $\lim _{n\to \infty }\left(\frac{n^2}{n^2}\cdot \frac{2}{5+\frac{2}{n}+\frac{1}{n^2}}\right)\:$  having the $n^2$ cancel  $\lim _{n\to \infty }\frac{2}{5+\frac{2}{n}+\frac{1}{n^2}}$ after taking the limit  $\frac{2}{5+0+0}$  so $\frac{2}{5}$$\ne $ $\:0\:$ so it's divergent?
19015
$$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$  Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write,   $$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$  $$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$  $$\log\left(2n\right) - \log\left(n\right) = \log(2)$$  Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor.
18688
Problem:  Show that $$\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}.$$  My progress:  The problem before this one had me find the Taylor series for $\ln(1-x)$ which was $$-\sum\limits_{n=1}^\infty \frac{x^n}n$$  so I figured I'd use $x=-1$ and plug that into the Taylor series. However, there was a side note stating that the Taylor series I found is only valid for $x\in(-1, 1)$. And in any case, my calculation isn't going anywhere, since I end up with the series $1-\frac12+\frac13-\frac14\cdots$  Question 1: How can I use this to solve the problem stated initially, and in the title?  Question 2: I can see why $x$ is restricted to be less than 1, to prevent taking the log of zero or a negative. Why is it not valid for x greater than 1?
116032
For what real values of $a$ does the series $$\sum_{n=1}^{\infty}\left(\frac{an}{n +1}\right)^n$$ converge?  At first, I wanted to do the root test, but the problem is that $a$ could be negative.  If I take it out of the fraction to get $$\left(\frac{a}{n+1}\right)^n$$ it doesn't meet the criteria of either the Dirichlet or the Abel tests.
124283
Show that $$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}\right)=\log 6$$ Here I need to use the definition of integral but I  faced problem in range . Please help.