Analyze the convergence of $$\sum\frac{1}{n(\log(n))^c}.$$  I know that it diverges for $c\leq1$ by using comparision test and integral test for convergence through the $$\int_{2}^{\infty}\frac{1}{x\log x}dx= \log(\log(\infty))-\log(\log2)$$  I also have a gut feeling that it converges for $c&gt;1$ but not able to prove so. Please try explaining through less exotic methods by not resorting to weird convergence test theorem. Thanks!
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).
36879
Does the following series converge:  $$\sum_{n=3}^\infty \frac {1}{(\log n)^{\log(\log(n)}}$$  I've tried all test I know... Any ideas ?
2411
Can someone give a simple explanation as to why the       $$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$   doesn't converge, on the other hand it grows very slowly?   I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.
80992
I tried to demonstrate the convergence of the following series by root and ratio test but they failed.  $$\sum_{n=1}^{\infty}\frac{\log(n)}{n^2}$$  Then I used Wolfram Alpha, which answered "By the comparison test, the series converges".  How can I do it?
61361
Could I have a hint for testing the convergence of the following series please?  $$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$  Edit  The integral test does not work because $\int_1^n\frac{1}{(\ln x)^{\ln x}}dx$ has not an elementary primitive.  Thank You.
39848
Does the following series converge or diverge:  $$\sum_{n=1}^\infty \frac{3^{1/\sqrt{n}}-1}{n}$$  Many thanks!
58869
I'm using the integral test to determine if this series converges. From what I have so far it seems that it diverges, but according to wolfram alpha it converges. Where is my mistake?   $$\sum_{k=2}^\infty {\frac{1}{k^2-1}} \to \int_{2}^M \frac{1}{x^2-1}dx$$  $$\int_{2}^M \frac{1}{x^2-1}dx = \frac{1}{2}\Big(\ln|M-1|-\ln|M+1|+\ln|3|\Big)$$  $$ \lim_{M\to \infty} \frac{1}{2}\Big(\ln|M-1|-\ln|M+1|+\ln|3|\Big)$$
47091
My question is pretty simple.  Since $n! \gt n!!$, it's clear by the comparison test that $\sum_{n=0}^\infty \frac {1}{n!!}$ converges.      But what value does the sum converge to? How does one go about determining its value (if possible)?
111930
While checking the convergence of the series $\sum u_n, u_n = \frac{n^n x^n}{n!}$ for $x&gt;0$, we used ratio test to say that for $0&lt; x &lt; \frac1e$ $\sum u_n$ is convergent and for $\frac1e &lt; x&lt;\infty$ $\sum u_n$ is divergent.  We use Logarithimic Test for the case $x = \frac1e$, where we came across computing the limit $$\lim_{n \to \infty} n+n^2 \log \frac{n}{n+1}$$ but I am stuck in finding the limit.
15598
Classify  $$\sum_{n=3}^\infty \frac{1}{n\log(n)}$$ as absolutely convergent, conditionally convergent or divergent.  Is it, $$\sum_{n=3}^\infty \frac{1}n$$ is a divergent $p$-series as $p=1$, and $$\lim_{n\to\infty} \frac{1}{n\log(n)}{n} = 0$$ by comparison test. And this converges to $0$.  So, $$\sum_{n=3}^\infty \frac{1}{n\log (n)}$$ is conditionally convergence?  I'm not sure if I'm doing right or not. Could you guide me?  Thanks in advance! :)
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?
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 $$\frac{\ln(n)+5}{n^2}$$
7191
Discuss the convergence of the sum:   $$\sum _{k=2} ^\infty \frac{(-1)^k}{\sqrt{k}+(-1)^k}.$$  My answer so far: It does not converge absolutely since   $$\left | \frac{(-1)^k}{\sqrt{k}+(-1)^k} \right | \geq \frac {1} {\sqrt{k} + 1},$$  the sum of whose terms diverges. Intuitively, I think it should converge conditionally because $ \sum \frac{(-1)^k}{\sqrt{k}}$ does, and the $(-1)^k$ in the denominator should not make a big difference. However, I am having trouble demonstrating its conditional convergence.  Evidently, the problem is that $\frac{1}{\sqrt{k}+(-1)^k}$ is not monotone. Let's call this sequence $a_n$. If we can show that   $$\sum_{k=2}^\infty |a_k - a_{k-1}| &lt; \infty,$$   then we're done by Dirichlet's test. But this seems to be $O(1/k)$.   Any ideas?
34891
I need to find out whether the following series converges or diverges. If it converges I also have to discover whether it is an absolute convergence or a conditional one. $$\sum_{k=1}^{\infty}\left(\frac{1+\cos k}{2+\cos k}\right)^{2k}$$ Root test did not help me much since $\lim_{k\rightarrow\infty}\left(\frac{1+\cos k}{2+\cos k}\right)^2$ may equal anything from $0$ to $4$. Comparison tests also did not work well. I guess this series is convergent, though. But I still have no idea how to prove it.
23637
I am trying to find out if the following sequence $(a_n)_{n\geq 1}$ will either converge or diverge. $$ a_n = (-1)^n + \frac1n $$  If the limit converges, I need to find its limit.   I tried plugging in various numbers for n besides 1, but I am not entirely sure if it really does converge.
47988
In my integration adventures, I came across this sum which I could not simplify:  $$\sum_{n=1}^{\infty}\frac{(-1)^{n}\log(2n+1)}{2n+1}$$   seems to believe the sum diverges and is not of much help here.  Does a closed form for this sum exist?  If not, can this sum be transformed nicely that has faster convergence?
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.