Evaluate if the following series is convergent or divergent: $\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\ln\ln{n}}$.   I cannot properly understand the notation that the book employs here $\ln\ln(x)$, but I guess it is referring to $\ln(\ln(x))$. Assuming $\ln \ln(x)=\ln(\ln(x))$,  I used the Weierstrass's or comparasion test to evaluate the series:  $$\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\ln\ln{n}}&lt;\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\ln(n)}=\sum\limits_{n=2}^\infty \frac {1} {n\ln^2{n}}&lt;\sum_\limits{n=2}^{\infty}\frac{1}{n^2},$$ proving the series converge. However the solution point out the series diverge.   I have already proved the convergence $\sum_\limits{n=2}^{\infty}\frac{1}{n^2}$.  Question:  What am I doing wrong? How can I prove the series diverge? Is $\ln\ln(x)=\ln(\ln(x))$ meant by the author of the book?
27188
I'm being asked to determine if $\displaystyle\sum\limits_{n=3}^\infty \frac1{n (\ln n)\ln(\ln n)}$ converges. So, using Cauchy's Condensation Test, I reduced the problem to one of determining the convergence of $\displaystyle\sum\limits_{n=3}^\infty\frac 1{n\ln (n\ln 2)}$. Am I on the right path, and how do I proceed from here?
25372
How to test the convergence of following series $$\sum{\frac{1}{(\ln {n})^{\ln{n}}}}$$  I have tried Cauchy condensation test and gives me nothing
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!
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.
117700
Determine whether the series $\sum_\limits{k=1}^∞ \frac2{k^{3/2}}$ converges or diverges. If it converges, find the bounds for the sum of the series. A) Converges, $\frac3{2}$ &lt; $\sum_\limits{k=1}^∞ \frac2{k^{3/2}}$ &lt; $\frac5{2}$ B) Converges, 2 &lt; $\sum_\limits{k=1}^∞ \frac2{k^{3/2}}$ &lt; 3 C) Converges, 4 &lt; $\sum_\limits{k=1}^∞ \frac2{k^{3/2}}$ &lt; 6 D) Converges, 6 &lt; $\sum_\limits{k=1}^∞ \frac2{k^{3/2}}$ &lt; 8 E) The series diverges I know the series converges. Using p-Series, p=$\frac3{2}$ meaning that p&gt;1 therefore the series converges. However, I don't know how to find the bounds for the sum of the series
29480
I would like to check if the series $\sum_{m=0}^{\infty}  {2m\choose m} \frac{1}{4^{m}} $ converges or diverges. The ratio test is inconclusive. Any other way to proceed? Thanks.
1163
Does this series converge or diverge? $$\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})$$   my thought is that, I can break it down to $$\sum_{k=1}^{\infty}\ln(k) - \sum_{k=1}^{\infty}\ln(k+1)$$ then maybe using comparison test or something? But I don't know exactly how to prove whether this series converges or diverges.  Any help would be appreciated!
34647
Does the following series converge or diverge? I would like to see a demonstration.  $$ \sum_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}}. $$  I can see that: $$ \sum_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}} \leqslant \sum_{n=1}^{\infty} \frac{1}{n^{1 + {\sin^{2}}(n)}} \leqslant \sum_{n=1}^{\infty} \frac{1}{n^{1 + {\sin^{2n}}(n)}}. $$
30738
How tipe are this series?  $$\sum_{n=2}^\infty \frac 1 {n^2\ln n}$$ But $$\sum_{n=2}^\infty \frac 1 {n\ln n\ln(\ln n)}$$ I used Condensation Test, but I stack after that. Thanks!
111999
How to evaluate this series for $k &gt; 1$? $$\sum_{n=1}^{\infty}\frac{1}{n k^n}$$  For $k = 2$, I tried to evaluate $\displaystyle \sum_{n = 0}^\infty  \int_{1}^{2} x^{-(n+1)}dx =   \int_{1}^{2} \sum_{n = 0}^\infty x^{-(n+1)}dx = \int_1^{2}\frac{1}{x(x-1)}dx$ $\displaystyle = \int_{1}^{2}\frac{1}{x-1}dx - \int_{1}^{2}\frac 1 x dx = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n2^n}\right)$  But both $\displaystyle \int_{1}^{2}\frac{1}{x-1}dx$ and $\displaystyle \sum_{n=1}^\infty\frac1n$ diverges. The answer is $\ln 2$, are these divergent terms equal?
89687
Does $\displaystyle\sum^\infty_{n=1} \ln(\frac {2^n+1}{2^n})=\sum^\infty_{n=1} \ln(1+\frac 1 {2^n})$ converge?   Condensation, root, ratio and limit comparison tests don't help, I don't know what to do... Any hints please?  Note: no integral test.
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?
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?
57252
How to prove that the series $$\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$$ is convergent? What about finding the sum?  My attempt: $$\ln (1-1/n^2)= \ln(n-1) -2\ln n + \ln(n+1)$$ The term in the log is decreasing so I get the feeling that the series converges but confused how to prove this and also find the sum.
68121
Does the following series converge or diverge?  $\sum_{n=1}^{\infty} a_n$ where  $a_n = \frac 1 b_n$, and $(b_n)_n$ is the subsequence of $(n)_n$ whose terms do not have a $9$ in their decimal notation.  I have no idea how the sequence $b_n$ even looks like. If I have an idea of how to prove that a series converges or not with criteria but I don't if I don't know what $b_n$ is.
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
18688
Problem:  Show that $$\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}.$$  My progress:  The problem before this one had me find the Taylor series for $\ln(1-x)$ which was $$-\sum\limits_{n=1}^\infty \frac{x^n}n$$  so I figured I'd use $x=-1$ and plug that into the Taylor series. However, there was a side note stating that the Taylor series I found is only valid for $x\in(-1, 1)$. And in any case, my calculation isn't going anywhere, since I end up with the series $1-\frac12+\frac13-\frac14\cdots$  Question 1: How can I use this to solve the problem stated initially, and in the title?  Question 2: I can see why $x$ is restricted to be less than 1, to prevent taking the log of zero or a negative. Why is it not valid for x greater than 1?