How can I prove that $\sum_{n=1}^{\infty} \frac{|\sin n|}{n}$ and $\sum_{n=1}^{\infty} \frac{\sin^2 n}{n}$ both diverge?  I thought of using Comparison test, but I couldn't find any sequence to compare with.  This question is from the book 'Real Analysis and Foundations' by Steven G. Krantz.
19379
It is stated as a problem in Spivak's Calculus and I can't wrap my head around it.
70361
Prove that $\displaystyle\sum_{n=2}^\infty\frac {(-1)^n}{\sqrt{n}+(-1)^n} $ diverges.  It is easy to see that this absolutely diverges, however how can it be proven to diverge in general? The idea as posted in another forum was to group the terms in the sequence as follows:  $$\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}+(-1)^n}=-\sum_{n=1}^\infty \left(\frac{1}{\sqrt{2n+1}-1}-\frac 1 {\sqrt{2n}+1}\right)$$  and then show that this sequence diverges. I did end up getting an answer from that forum (thank you!) however it depended on results from WolframAlpha which seem extremely difficult to solve by hand. This question is from Apostol's Calculus, written in 1969, so I'd like to have a solution which doesn't depend on WolframAlpha.
88110
I am given the problem: Let $a_n&gt;0$, prove $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \sin(a_n)$ converges.  We have this problem in a homework, and I don't believe it can be true. The statement being biconditional implies that when $\sum_{n=1}^{\infty} \sin(a_n)$ converges, so too does $\sum_{n=1}^\infty a_n$, but if we take $a_n=\pi n$ or even $a_n=\pi$, then $\sum_{n=1}^\infty \sin(a_n)$ converges while $\sum_{n=1}^\infty a_n$ diverges. Is my thinking incorrect?
15494
Determine whether the series $\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}$ is convergent or divergent.  I've tried applying all the basic tests to no avail. I need to find out what the "trick" for this particular series is.
38229
Given the series $S=\sum_{n=1}^{\infty} \frac{(2n)!}{4^n(n!)^2}$ I'm trying to prove its divergent but with no luck. It doesn't tend to $\infty$ as $n$ grows large and ratio test is inconclusive. I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?  Thanks!
88825
$\sum_{n=1}^\infty n*(n+1)*x^n$  Hello everyone,  I need help in solving the question above.  I started with $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, if differentiated it once so it became $\frac{1}{(1-x)^2} =\sum_{n=1}^\infty nx^{n-1} $ but from here I don't know how to continue. Thanks.
39045
Is the series  $$ \sum_{n=1}^\infty \frac{|\sin n|^n}n\tag{1}$$  convergent？  If one want to use Abel's test, is   $$ \sum_{n=1}^\infty |\sin n|^n\tag{2}$$  convergent？  Thank you very much
23900
Per the title, if $\sum_{1}^{\infty}(a_n)^3$ diverges, does this imply that $\sum_{1}^{\infty}(a_n)$ diverges?   I'd appreciate hints (!) for dealing with this excercise.  EDIT Per the contrapositive, it is not given that $a_n$ converges absolutely, or that it is nonnegative for all $n$.  Thank you!
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
1983
Let $a_1,a_2,a_3,\ldots$ be reals. Prove that if $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$.   For this I have shown the case for when $ a_n^{2} \le\frac {|a_n|}{n}$ $\Rightarrow$ $ |a_n|\le\frac {1}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le \frac{1}{n^{2}}$ and we know that $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges and hence $\sum_{n=1}^{\infty}\frac {a_n}{n}$ converges by the comparison test. Now considering $ a_n^{2} \ge\frac {|a_n|}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le a_n^{2}$ $\rightarrow$ combining the two cases for any n we have: $\frac {|a_n|}{n}\le\frac{1}{n^{2}}+a_n^{2}$ Hence using the comparion test again we know that $\sum_{n=1}^{\infty} a_n^{2}$ converges and $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges hence the sum converges so we can conclude that $\sum_{n=1}^{\infty} \frac {a_n}{n}$ is absoluetly convergent $\Rightarrow$ convergent.  Not to sure if this is correct, any help would be much appreciated, many thanks.
91354
Show that $\sum_{k=1}^\infty \frac{\ln{k}}{k^2}$ converges or diverges   I know that I must use the Limit Comparison test and my instinct tells me that this series will converge. I cannot, however, think of another series $b_k$, to compare my given series to in order to show that it converges.  Any hints?   Thank you!
69560
Prove that if $a_n\gt 0$ and $\sum a_n$ diverges, then $\sum \frac{a_n}{1+a_n}$ diverges.     This is the solution to this problem, but I'm having a hard time understanding it. Why does $a_k/(1+a_k)$ not converge to $0$ if $a_k$ doesn't converge to $0$?   I'd appreciate it if anyone could answer this question for me.
55725
I'm struggling to prove that $$\sum \limits_{n=1}^{\infty} \frac{\cos(\log(n))}{n}$$ diverges.   Does anyone have any idea how to prove it? Breaking it into smaller pieces din't work. Maybe I should bound it with another series? But how?
1102
I need to prove the following:  $$\lim_{s \to \infty} \sum_{n = 1}^\infty \frac 1 {n^s} = 1$$  This is my attempt:  \begin{align} \lim_{s \to \infty} \sum_{n = 1}^\infty \frac 1 {n^s} &amp; =  \lim_{s \to \infty} \left( 1 + \sum_{n = 2}^\infty \frac 1 {n^s} \right) \\ &amp; = 1 + \sum_{n = 2}^\infty \lim_{s \to \infty} \frac 1 {n^s} \\ &amp; =  1 + \sum_{n = 2}^\infty 0 \\ &amp; = 1 \end{align}  I need help justifying the interchange of the sum and the limit though. Is it enough to show the limit exists and the sum is convergent? I read that I need to prove uniform convergence, how can I do this?
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?
65756
As far as I know, the textbook approach to determining the convergence of series like $$\sum_{n=1}^\infty\frac{\cos n}{n}$$ and $$\sum_{n=1}^\infty\frac{\sin n}{n}$$ uses Dirichlet's test, which involves bounding the partial sums of the cosine or sine terms. I have two questions:   Are there any other approaches to seeing that these series are convergent? I'm mostly just interested to see what other kinds of arguments might be made. What's the best way to show that these two series are only conditionally convergent? I don't even know the textbook approach to that question.
104695
Could you tell me how to prove that if two series $ \sum_{n=0} ^{\infty}x_n, \sum_{n=0} ^{\infty} y_n$ are convergent, then $\sum_{n=0} ^{\infty}(\alpha \cdot x_n + \beta \cdot y_n)$ is also convergent and its sum equals $\alpha \cdot \sum_{n=0} ^{\infty} x_n + \beta \cdot \sum_{n=0} ^{\infty} y_n$?