I know that the series $\sum \limits _{n=1}^\infty (1- \cos\frac{1}{n})$ converges.   One way is to use trigonometric identity which says $(1- \cos\frac{1}{n})$ = $2\sin^2(\frac{1}{2n})$ and from here I need to show that $\sum \limits _{n=1}^\infty 2 \left(\frac{1}{2n}\right)^2 $ converges, which is pretty simple because the denominator's $n^p$ and $ p &gt; 1 $.  But I want to try another approach. I know that the beginning of $\sin(\frac{1}{n})$ Maclaurin series is ($x_0 = 0$)$ 1 - (\frac{1}{n})^2 \ 2! $. which means I need to check: $\sum \limits _{n=1}^\infty (1- (1 - (\frac{1}{n})^2 \ 2!))$ = $\sum \limits _{n=1}^\infty (\frac{1}{n})^2 \ 2!$  and this is clearly converges because it is harmonic sequence where $n^p$ and $ p &gt; 1 $.  Does it make sense what I just wrote?       What I struggle with: Is my Maclaurin evaluation true?
51854
I'm having trouble determining the convergence of the series:  $$ \sum_{n=1}^{\infty}\left[1-\cos\left(1 \over n\right)\right]. $$  I have tried the root test:  $$\lim_{n\rightarrow\infty}\sqrt[n]{1-\cos\frac{1}{n}}=\lim_{n\rightarrow\infty}\left(1-\cos\frac{1}{n}\right)^{1/n}=\lim_{n\rightarrow\infty}\mathrm{e}^{\frac{\log(1-\cos\frac{1}{n})}{n}}=\mathrm{e}^{\lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n})}{n}}$$  Now by applying the , that upper limit is equal to:  \begin{align} \lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})}{(n+1)-n}&amp;=\lim_{n\rightarrow\infty}\left(\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})\right) \\&amp;=\lim_{n\rightarrow\infty}\log{\frac{1-\cos{\frac{1}{n+1}}}{1-\cos{\frac{1}{n}}}} \end{align}  Now I'm totally stuck, unless that quotient is actually 1, in which case the limit would be 0, the Root test result would be $\mathrm{e}^0=1$ and all this would have been to no avail.  I'm not sure this method was the best idea, the series sure seems way simpler than that, so probably another method is more appropriate?
22074
The series $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n \log n}$ is known to diverge. This can be seen with comparison to the harmonic series.   The other series $\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{n \log n}$ converges due to Liebniz's alternating criterion. I see these two series quite often in exams' questions to examine them if they converge or not. The answers are pretty straight forward.   Now , out of curiosity , could we find a closed form for the series  $$\mathcal{S}=\sum_{n=2}^{\infty} \frac{(-1)^n}{n \log n}$$  The interesting fact is that if we define the function   $$f(s)=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^s \log n} , \; s \geq 1$$  then taking the derivative with respect to $s$ we get that   $$f'(s)=-\sum_{n=2}^{\infty} \frac{(-1)^n}{n^s} = \zeta(s) - 2^{1-s} \zeta(s) - 1$$  I don't know how to go back and evaluate $f(1)$. Has anyone done that before. Expressing the value of the series in terms of special functions will suit me. Numerically it is approximately equal to $0.526412$. That said Wolfram. So, any clues?
123584
Sorry for the oversimplified question, but does the series $\sum_{n=1}^{\infty} \frac{1}{n+1}$ converge? The ratio test of it gives the result of "1". Thanks a lot.
45732
Does this series converge or diverge?  $$\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt{n}}{n}$$  I tried the comparsion test with $\sum_{n=1}^\infty$ $1 \over n$ but this does not help.  I also tried to simplify the series to $\sum_{n=1}^\infty \frac{1}{n (\sqrt{n+1}+\sqrt{n}{}{}{})}$ but this become harder.
1602
How can I evaluate $$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$? I know the answer thanks to , but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.  In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
118828
$$\sum_{n=1}^\infty \frac{1}{n^2 2^n}$$  I am new in series thus I tried a pair of methods to compute but I couldn't
12372
Does the following series converge? $$\sum_{n=1}^\infty \frac{1}{n\sqrt[n]{n}}$$   As $$\frac{1}{n\sqrt[n]{n}}=\frac{1}{n^{1+\frac{1}{n}}},$$ I was thinking that you may consider this as a p-series with $p&gt;1$. But I'm not sure if this is correct, as with p-series, p is a fixed number, right ? On the other hand, $1+\frac{1}{n}&gt;1$ for all $n$. Any hints ?
2819
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$   I found that:  $$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$  and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}$ but the answer is $\frac{3}{4}$
62807
I tried everything that I know but I couldn't solve this series:  $$\sum_{n=0}^{\infty} \left(\frac{\pi}{2} - \arctan(n)\right)$$  Does it diverge or converge?
98662
At  I've read that being $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ by using  we get $\sin'(x)=\sum_{n=0}^{\infty}(2n+1) \frac{(-1)^n}{(2n+1)!}x^{2n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=\cos(x)$ but the page  says that if $f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$ then inside the radius of convergence of the power series we have $f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1}$ so shouldn't it be $\sin'(x)=\sum_{n=1}^{\infty}(2n+1)\frac{(-1)^n}{(2n+1)!}x^{2n+1-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$?  Now I understand that this last result has a missing term due to the index of the series being $1$ and not $0$ but being the $1$ mandatory because of the above theorem and since trying to shift index doesn't help (we get $\sin'(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2}$) how can the result that ProofWiki shows be rigorously justified?   (I mean, I think we can't just say "oh, there's a missing term!" after having applied the theorem and magically shift the index of the series to obtain the desired result)
86270
We know that $\sum \frac{1}{n^p}$ is convergent for $p&gt;1$. However the series $\sum {\frac{1}{n^{1+1/n}}}$ is apparently divergent since $1+1/n$ tends to 1 as $n$ tends to infinity. But how to prove this? The root test fails expectedly and I haven't been able to find a smaller divergent series for comparison test. Can someone help? Thanks!
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?
59042
I am trying to determine if the sum $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ is convergent or divergent.   In attempting to solve this problem I began by seeing if I could do the divergence test - in this case it is not helpful. The terms $4^n,6^n$ look like geometric series type terms but the $n$ in the numerator and denominator are troublesome. I first tried factoring $n$ out of the expression:  $$\sum_{n=1}^\infty \frac{n+4^n}{n+6^n} = \sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$$  Next I tried comparing to the series $$\sum_{n=1}^\infty \frac{\frac{4^n}{n}}{\frac{6^n}{n}} = \sum_{n=1}^\infty \frac{{4^n}}{{6^n}} = \sum_{n=1}^\infty{\left(\frac{2}{3}\right)}^n$$  This is a convergent geometric series as $r = \frac{2}{3}\lt1$  now I reason that the series $\sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$ is neither smaller not greater than the series $\sum_{n=1}^\infty{(\frac{2}{3})}^n$ Am I able to conclude anything in this case by a direct comparison test or can I only conclude that the first series is convergent if the second is greater than the first?  Thanks
47326
How do you prove $$\sum_{n=1}^{\infty} \frac {n}{2^n} = 2\ ?$$  My attempt: I have been trying to find geometric series that converge to 2 which can bind the given series on either side. But I am unable to find these. Is there a general technique to find the sum? This is a high school interview question and must be easy enough to solve in a few minutes.  Please give any hints for the first step towards a solution.
33274
The sum is $$\sum_{n=1}^{\infty} \frac{n+2^n}{n+3^n}$$ Is this convergent or divergent? I tried to use the divergent test but the test fail because $a_n = (n+2^n)/(n+3^n) = 0 $ as $n$ goes to infinity. Could someone point me to the right direction?  Thanks
19946
If I have $\sum_{n=2}^{\infty} \frac {1}{n\log n}$ and want to prove that it diverges, can I use following?  $$\frac {1}{n\log n} \lt \frac {1}{n}$$   $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of $\frac 1 n$ equals to zero so the comparasion I think isn't useful. Or can I say it diverges because $\sum_{n=1}^\infty \frac 1{\log n}$ also diverges?   And If I have $\sum_{n=2}^\infty \frac {(-1)^n}{(n)}$ I can use the Leibniz rule. $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of it equals to zero, so I am confused if this series diverges/converges again. I understand that $\sum_{n=1}^\infty \frac 1 n$ has an infinite sum, but what to do in this case?
53658
Consider the series:  $$a) \sum_{n=1}^\infty \frac{a^n}{ \prod_{k=1}^n \ (1+\sin\frac{1}{2k})}$$ $$b) \sum_{n=1}^\infty \frac{a^n}{ \prod_{k=1}^n \ (1+\tan\frac{3}{2k})}$$  Showing that these two are convergent (and absolutely convergent) it's no big deal for  $a\in R^*$. But I couldn't figure out how to solve it when $a=1$. My guess is that the general term has to be brought to another form. And that is my question: how to prove convergence when $a=1$.  Thanks for helping with this!