I have to study the character of this series  $$\sum_{n=1}^\infty \frac{1}{n^{\alpha}(\log  n)^{\beta}}.$$  $\alpha$ and $\beta$ are two parameters.  I'm considering the case $\alpha &gt;1 $ and all real values for $\beta $ .  If $\beta \ge 0$, $\exists n_0 \in N $ such that $\forall n\ge n_0, \dfrac{1}{n^ {\alpha}(\log  n)^ {\beta}}&lt;\dfrac{1}{n^ {\alpha}}$.   For the comparison test $\dfrac{1}{n^{\alpha} (\log  n)^ {\beta}}$ converges.  I tried to analyze the case $-1&lt;\beta&lt;0,\beta=-1, \beta&lt;-1$ but I'm not sure.  Anyway, for $\beta&lt;-1$ the series is convergent for $\alpha &gt; 1- \beta$, divergent for $\alpha &lt; 1- \beta$.  For $-1&lt;\beta&lt;0$ the series is convergent for $\alpha &gt; 2$, divergent for $\alpha &lt;2$.  For $\beta=-1$ the series is convergent for $\alpha &gt; 2$, divergent for $1&lt;\alpha &lt; 2$.  I've used the fact that $\log n&lt;n$
37468
Study the convergence of $$\sum_{n=2}^{\infty}  \frac{1}{n^{\alpha}\ln^{\beta}(n)}$$ where $\alpha, \beta \in  \mathbb{R}$   I have proved that:   This series diverges when $\alpha \leq 0$. This series converges when $\alpha &gt; 1, \beta &gt; 0$ This series diverges when $0 &lt; \alpha &lt; 1, \beta &gt; 0$ This series converges when $\alpha = 1, \beta &gt; 1$      Question: What happens when $\alpha &gt; 0$ and $ \beta &lt; 0$?   There are other questions on MSE which ask about this series, but this question is distinct because   I would like an argument which does not rely on the integral test for series convergence, and this question considers all real $\alpha$ and $\beta$, while other questions ask only about $\alpha, \beta &gt; 0$, where we can apply Cauchy condensation criterion
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.
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
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.
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).
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?   $$\sum_{k=2}^\infty {\frac{1}{k^2-1}} \to \int_{2}^M \frac{1}{x^2-1}dx$$  $$\int_{2}^M \frac{1}{x^2-1}dx = \frac{1}{2}\Big(\ln|M-1|-\ln|M+1|+\ln|3|\Big)$$  $$ \lim_{M\to \infty} \frac{1}{2}\Big(\ln|M-1|-\ln|M+1|+\ln|3|\Big)$$
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?
68289
Question is to check if : $\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}&gt;\frac{3}{5}$   the problem is that i am sure that this series$\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}$ is convergent (if it is not then i would be very happy as i will then be done.)  I would be thankful if someone can help me out with solving this.
56578
$$ \sum_{i=0}^\infty \frac{(-1)^n}{n} $$  This alternating series fails the p-series test because the exponent of n = 1.  Yet it seems to pass the alternating series test.   1 - $a_n$ must be positive. True.  2 - Terms must be decreasing. $\frac{d}{dn} 1/n = -n^{-2}$, which is &lt; 1. True.  3 - $ \lim_{n\rightarrow\infty} 1/n = 0  $ True.  $(-1)^n/n$ is clearly a divergent series, so why does it pass the AST?
119969
Suppose $\sum_{n=2}^{\infty} a_n$ is a convergent series. Is the following series convergent as well? $$\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)$$ The part with $n^{a_n}-1 $ looks quite similar to $\sqrt[n]{n}-1$ which behaves like $\ln{n}$, however, I'm not able to tie this fact to the behavior of the whole expression. Any suggestions?
60070
How can we evaluate this series  $$\sum_{n=1}^\infty \frac{n^a}{b^n}?$$  Here $a$ and $b$ are positive integers. If $b=1$ then series will be diverging, in other cases, it will be converging, but how to find this sum?  Isn't there a simple solution which doesn't involve Stirling numbers?
30910
Could I have a hint for testing the convergence of the following series please?  $$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$  I am very appreciative for your help.
68131
I recently encountered the following problem: If $\sum a_n^2 $ converges and $\alpha_n= \frac{a_1+...+a_n}{n}$ then show that:  $$4\sum_{n \ge 1} a_n^2 \ge \sum_{n \ge 1} \alpha_n^2$$  I had an approach which i did not manage to utilize, so i am asking if anyone has an idea if it can be used or not:   We note that the right hand side will be maximized if $a_i$ are all of the same sign. Without Loss of Generality, suppose that $a_i \ge 0$. If we use (1.), another way to maximize the right hand side would be if $a_i \ge a_{j}$ for all $i&gt;j$. Let us also suppose that this is true.   And i stop there, i tried many approaches but nothing seemed to work. I thank you in advance for any hint and/or a complete solution.  Supplementary question: Can we construct a similar problem for integrals (instead of series)?
118828
$$\sum_{n=1}^\infty \frac{1}{n^2 2^n}$$  I am new in series thus I tried a pair of methods to compute but I couldn't
104617
Wolfram alpha stated that this series converges to $0,$ but when I entered this answer in my homework it says it is incorrect, therefore I wanted to know whether this series diverges or converges.
27086
$$\sum_{n=1}^\infty \frac{i^n}{n}$$  I'm not sure how to approach this series to find if it converges/diverges because of the $i^n$. I tried using the comparison test comparing it with $\frac{1}{n}$ and the ratio test, but didn't get anywhere. How would you approach this and look at series of this form or how to look at $i^n$ in an infinite series?
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?