Two questions:  Show whether $\sum_{n=2}^{\infty} \frac{1}{n \ln(n^3)}$ converges or diverges.  According to wolfram, the series diverges by the comparison test, so I tried the following:  for $n$ large, we get $n &lt; n \ln(n^3) $ but then $\frac{1}{n} &gt; \frac{1}{n \ln(n^3)}$, which gives me nothing, any tips?
122651
Is $\sum_{n=2}^\infty (n\ln n)^{-1}=\infty$ ?  This seems like elementary calculus, but I can't figure this out.  Can anyone supply a hint?
115877
Test for convergence  $$\sum_{n=1}^{\infty} \frac{1}{n^n}$$  I'm at a loss on what to do, is this a geometric series $\frac{1^n}{n^n}$?
21194
So my teacher gave a pretty tough (for me) problem in class today: Does the series $$ \sum_{n=1}^{\infty}\frac {\sin{\frac 1 n}} {\sqrt n} $$ converge or diverge?  So far I've thought of trying the Comparison Test and Ratio Test but couldn't really get anywhere. Intuitively I think it should converge, but how do I actually show 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?
34849
I'm trying to show that the series $\sum_{n=1}^\infty\frac{n}{n^2+1}$ is divergent.   I'm trying to use a comparison test to do so. To do so, I first can see that $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{n=1}^\infty\frac{n}{n^2+n} = \sum_{n=1}^\infty\frac{1}{n+1}$ And I can use a change of variable $k = n+1$, so show $\sum_{n=1}^\infty\frac{1}{n+1} = \sum_{k=2}^\infty\frac{1}{k} = \infty$. Then since $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{k=2}^\infty\frac{1}{k}$ it must be that $\sum_{k=2}^\infty\frac{1}{k}$ also diverges to infinity. Is this a valid way to show it?
99381
I'm trying to prove this sequence converges:  $\sum_{n=1}^{\infty}  \frac{1}{n\ln^2(n)}$  I noticed that this is continuous function which its derivative is always less than $0$ for $ x \gt 1 $, so I tried to do the integral test but with no success...   Some help?  Thanks!
81247
Does the following series converge or diverge? $$ \sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}} $$ The methods I have at my disposal are geometric and harmonic series, comparison test, limit comparison test, and the ratio test.
43842
Let $\sum a_n$ be a series of non-negative terms and let $$L = \lim_{n\to\infty}n\left(1-\frac{a_{n+1}}{a_n}\right)$$ Prove that the series converges (resp. diverges) if $L &gt; 1$ (resp. $L&lt;1$). I've tried, for example, that when $L&lt;1$, $$n\left(1-\frac{a_{n+1}}{a_n}-L\right)= \frac{n(a_n-a_{n+1}-La_n)}{a_n}\ge\frac{n(a_n-a_{n+1}-a_n)}{a_n}=\frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?  P.S. using Kummer's test doesn't count
83494
If $\sum_{n=2}^{\infty}a_n$ converge, then also converge this series? $$\sum_{n=2}^{\infty}\frac{\sqrt{a_{n}}}{\ln\, n}(n^{a_{n}}-1)$$  Please verify my answer below  Counterexample:  $$a_{n}=\begin{cases} \frac{1}{k^{2}} &amp; n=k!^{k^{2}}\\ \\ 0 &amp; \text{All other cases} \end{cases}$$  When our infinites sum is equal to  $$\sum_{k=1}^{\infty}\frac{1}{k}\cdot\frac{1}{2\, \ln\, k}\cdot(k!-1)$$
50464
Prove that $\displaystyle \sum_{n=1}^\infty \;\ln\left(n\sin\frac{1}{n}\right)$ converges.   My Work: $$\left|\ln \left(n \sin \frac{1}{n}\right)\right| \leq\left|\ln \left(n \sin \frac{1}{n^{2}}\right)\right| \leq\left|\ln \left(\sin \frac{1}{n^{2}}\right)\right|$$ I was going to use comparison test. But now stuck. Please give me a hint.
114346
I am trying to study the convergence of the series:  $$\sum_{n =2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$$ First I have failed to use the D'Alembert test  $$ \lim_{n\to \infty} \frac{(\ln (n+1))^{\ln (n+1)}}{(\ln n)^{\ln n}} =1.$$ So I can Conclude anything from here.  I have also tried the integral Comparaison test but it was not soo clear.
4665
Recently, I encountered a problem about infinite series. So my question is how to know whether the infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ is convergent?
114562
Decide if the series converges and prove it using comparison test:   $\sum_{k=1}^{\infty}\frac{1}{2k+3}$   $$\frac{1}{2k+3} &lt;\frac{1}{2k}&lt;\frac{1}{k}$$  and since $\sum_{k=1}^{\infty}\frac{1}{k}$ diverges (we have defined that in our readings, so I can claim it here), the complete series will diverge, too.  Did I do everything correctly? What makes me feel a bit unsure is that there is no "$\leq$" sign anywhere and when I look up the comparison test on the internet, I don't see any "&lt;" or ">" in the definition.  Is it correct anyway?
5047
I'm not completely new to convergent series and sequences, but proving the above gave me some troubles. Would appreciate some help :) Show that: $$\lim_{n\to\infty} \frac{\sum\limits^n_{k=1} \frac{1}{k}}{ \ln(n) } = 1$$
35907
Hello I am a high school student from germany and I am starting to study math this october. I am trying to prepare myself for the analysis class which I will attend so I got some analysis problems from my older cousin who also studied maths. But I am stuck on this problem. Check the following series for convergence/divergence $$\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$$ I tried to prove the convergence by comparison test  $$\sqrt{n^3+1}-\sqrt{n^3-1}= \frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{1}{n^2} \cdot \frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}$$ and then compare it with $$\sum \limits_{n=1}^{\infty}\frac{1}{n^2}$$  But in order to do that, I need to prove that $$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1$$ But I am having problems to prove that. Does anyone have tip how to solve this problem?
50422
I have $\displaystyle\sum_{n=1}^{\infty}{\frac{(2n)!}{(4^n)(n!)^2(n^2)}}$ and need to show whether it diverges or converges.   I attempted to use the ratio test, but derived that the limit of $\dfrac{a_{n+1}}{a_n}=1$ and hence the test is inconclusive.   So I now must attempt to use the comparison test, but I am struggling to find bounds to compare it to to show either divergence or convergence.
108084
Determine whether the following series is convergent or divergent  $$\sum_{n=0}^∞ \frac{(\ln(n)+5)}{n^2}$$  I tried using comparison test and p-test: $$\frac{\ln(n)+5}{n^2}$$&lt;=$$\frac{1}{n^2}$$  And since 1/n^2 is convergent by p-test, the series is convergent. But i am not sure if my solution is correct especially the part on determining the upper bound of $$\frac{\ln(n)+5}{n^2}$$