I can fairly easily work out that $\sum (\log k)^{-k}$ does converge using the ratio test, and I think I could use that somehow to work out whether $\sum (\log k)^{- \log k}$ converges. The ratio test and the root test don't give any result on this series, so I think there might be a clever comparison test I could do, but I don't see it.
84138
As the title suggests, I'd like to prove that the sum $$ \sum_{n=2}^\infty (\ln n)^{- \ln n} $$ is finite.  The root and ratio test both fail here, but  suggests that there is a comparison that can be used to show convergence.  The only thought I have is that it may help to write the terms as $e^{-\ln(n)\ln(\ln(n))}$, but this has not led me to any particular insight.  Any ideas are appreciated.
95762
As a follow-up to a , I'd like to know if  $$\sum_{n=1}^\infty\frac{1}{2 ^ n}$$  is convergent? If yes, then what does it converge to?  Just as a sidenote - the whole issue started with reading about Zeno's Turtle Paradox which made me curious about the underlying mathematics.
88888
My attempt runs as follows. Since $\sum |a_{n}|$ converges, the series $\sum a_{n}$ converges. Since $a_{n}^{2} \geq 0,$ it suffices to prove that there is an $M &gt; 0$ such that $$\sum_{1}^{m}a_{n}^{2} &lt; M$$ for $m \geq 1.$ But, since $\sum a_{n}$ converges, there is a $B &gt; 0$ such that $$\sum_{1}^{m}a_{n} &lt; B$$ for $m \geq 1,$ so $$(\sum_{1}^{m}a_{n})^{2} = \sum_{1}^{m}a_{n}^{2} + 2\sum_{j &lt; k}a_{j}a_{k} &lt; B^{2}$$ for $m \geq 1.$  However, the problem is there seems no way to determine whether the sum $\sum_{j &lt; k}a_{j}a_{k}$ is positive or negative?
91606
I'm supposed to determine whether this sum diverges or converges and if it converges then find its value:  $$ \sum_{n=2}^\infty \frac{1}{n^2-1}. $$  Using the comparison test I eventually showed that this converges.  But I can't figure out how to show what this sum converges to.  The only sums we actually found values for in my notes are geometric series which this clearly isn't.  I saw that I could use partial fraction decomposition to represent the terms as  $$\frac{1/2}{n-1}- \frac{1/2}{n+1} $$ but that just gets me $\infty - \infty$, so this isn't the way to do it.  I'm not sure how to find the value of this sum.  I don't need the full solution but a hint would be appreciated. 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!
108692
This is the series $\sum_{k=1}^{\infty} \frac{7^{k} + k}{k! + 1}$. I began with the ratio test, but after I set up the ratio to take the limit of, I couldn't simplify the limit, since nothing would cancel out. I believe it converges, but I have no idea how to solve further after setting up that ratio test limit.  If someone has another method, that would be fine too. We are not allowed to use Comparison or Limit Comparison Tests.
55937
I have to decide whether the Series $\sum_{n=1}^\infty \frac{n^2}{2^n}$ Converge or Diverge only by using the Comparison test or by $\frac{1}{n^p}$ test.  What I noticed is that it converges by the De'lamber test ($\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$).  But I'm having trouble finding a way proving it with the provided tests.  Would appreciate your advice
30633
Does $\sum_{k=1}^\infty\frac1{k^n}$ converge for $\Re(n)=1,\Im(n)\ne0$?  The ratio test is inconclusive.  It passes the term test for $\Re(n)=1$, but this is not sufficient to prove convergence.  Since we are dealing with so many complex numbers, I do not know of any convergence tests for this.  I know that if it is convergent, it is conditionally convergent.
54459
We all know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}$ converges for $s>1$ and diverges for $s \leq 1$ (Assume $s \in \mathbb{R}$).  I was curious to see till what extent I can push the denominator so that it will still diverges.  So I took $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\log n}$ and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges).  No surprises here. I expected it to diverge since $\log n$ grows slowly than any power of $n$.   However, when I take $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\log n)^s}$, I find that it converges $\forall s>1$.  (By the same argument as previous).   This doesn't make sense to me though.   If this were to converge, then I should be able to find a $s_1 > 1$ such that  $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$  Doesn't this mean that in some sense $\log n$ grows faster than a power of $n$?   (or)  How should I make sense of (or) interpret this result?  (I am assuming that my convergence and divergence conclusions are right).
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.
53321
I'm trying  to find out now whether the series $\sum_{n=2}^{\infty } a_{n}$ converges or not when $$a_n = \frac{n^{\log n}}{(\log n)^{n}}$$  Again, I tried d'Alembert $\frac{a_{n+1}}{a_{n}}$,   Cauchy condensation test $\sum \limits_{n=2}^{\infty } 2^{n}a_{2^n}$, and they both didn't work for me.  I can't use Stirling, nor the integral test.  Edit: I'm searching for a solution which uses sequences theorems and doesn't involve functions.    Thank you
60987
We had the following theorem in class:     Let $(a_n)$ and $(b_n)$ be sequences and $b_n&gt;0$ and $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=L$ with $L\in\mathbb R\backslash\{0\}$. Then $\sum a_n$ converges if and only if $\sum b_n$ converges.   So by recapitulating the lecture I've tried to prove it but I didn't get it. It's obvious that you have to use the comparison test, but how? Can anybody help? Thanks a lot!
106773
Does $\sum\limits_{n=3}^{\infty} \dfrac{1}{(\log\log n)^{\log n}}$ converge or diverge ? I tried some tests , but nothing conclusive is coming . Pleas help
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.
48067
I can clearly see that $\dfrac{n!}{n^n}\to 0$ when $n\to\infty$. But how do I know if the sum $$\sum_{n=1}^{\infty}\frac{n!}{n^n}$$ is convergent or not? I know this might be basic, but thank you if anyone can help me.
54754
While revising for exams, I came across a question where at the end of it, we had to determine whether the below series was convergent or divergent:  $$\sum_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $$  Unfortunately, other than knowing that as the series $ \sum_{n=1}^\infty \frac{(-1)^n}{n} $ converges, this series will most likely converge as the $(2 + (-1)^n)$ terms are bounded, I can't see a way of using this to determine if it converges or not.  Here's a list of some of the other ideas I've tried, which don't seem to get me anywhere:   Generalising the series - By this, I mean replacing the $-1$ for $z$ and then try and use the ratio test to see whether $-1$ is inside or outside or on the boundary of the circle of convergence - however, when I did so, I believe that the ratio test is inconclusive for all values of $|z|$, which while disappointing, is an interesting feature of the series.  - While this is usually a nice tool to use for when dealing with awkward sums, I can't see any good choice of sequences $ (a_n) $ and $(b_n)$ to choose to use. Summing consecutive terms - By this, I mean evaluating the series by looking at the sum of    $$\sum_{k=1}^\infty \frac{(-1)^{2k}}{2k(2+(-1)^{2k})} + \frac{(-1)^{2k-1}}{(2k-1)(2+(-1)^{2k-1})}$$  $$ \Rightarrow \sum_{k=1}^\infty \frac{1}{2k} - \frac{1}{3(2k-1)}$$  $\quad$ but then as we are left with parts of the harmonic series this doesn't seem like a nice way to  $\quad$evaluate it.  Other than that, I'm completely out of ideas, so any new ones (or ways to make my old ones work) would be much appreciated!
91006
$$\sum_{n=2}^\infty \frac{1}{(\ln\, n)^2}$$  The series converge?  Please verify my solution below