Give an example of a series $\sum a_n$ such that $\sum a_n$ is convergent but $\sum a_{3n}$ is divergent.  Here an example by zhw. is given as $$\frac 1 2 + \frac 1 2 - \frac 2 2 + \frac 1   3 + \frac 1 3 − \frac 2 3 + \frac 1 4 + \frac 1 4 − \frac 2 4 + \frac 1 5 + \frac 1 5 − \frac 2 5+\cdots.$$   But I need a better result with functional expression.
100736
Give an example of a convergent series $$\sum_{n=1}^{\infty}a_n$$ such that the series $$\sum_{n=1}^{\infty}a_{3n}$$ is divergent..  I cannot find find any kind of series... I am also be very thankful if you find a divergent series (changing the term convergent)...
36236
Is there a (real) sequence such that $\sum{a_n}$ diverges but $\sum{na_n}$converges?
113842
Given the series $A=\sum\limits_{n=1}^{+\infty}\frac{p_n}{p_{n+1}}$ and $B=\sum\limits_{n=1}^{+\infty}\frac{p_{2n}}{p_{2n+1}}$, where $p_n$ is the sequence where the nth number are the nth prime number, is it possible to determine whether they are converging or diverging?   I think there are some ways using the fact that $\sum\limits_{p\in\mathbb{P}}\frac{1}{p}$ diverges, but I'm not sure of how I can use it.
67159
Suppose $\sum_{n=1}^\infty \frac{1}{a_n} = A$ is summable, with $a_n &gt; 0,$ $n = 1,2,3,\cdots.$ How can we prove that $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n}$ is also summable?  This question came from a problem-solving seminar, but I'm quite stuck without a push in the right direction. I tried a few things, including Cauchy-Schwarz (which says $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n} &lt; \sum_{n=1}^\infty \frac{A}{n}$) and also the idea of assuming the latter series diverges and attempting to deduce the divergence of the former series from that, using facts such as $\sum a_n = \infty \implies \sum \frac{a_n}{a_1 + \cdots + a_n} = \infty$. Nothing has worked so far.
74043
Does the series $$ \sum_{n=1}^\infty \frac{\sin^2(n)}{n} $$   converge?   I've tried to apply some tests, and I don't know how to bound the general term, so I must have missed something. Thanks in advance.
82734
It can be easily show that the harmonic series $$\sum_{n=1}^{\infty}\dfrac{1}{n}$$ is divergent. Also it has shown that the infinite series of reciprocals of primes $$\sum_{p\text{ is prime}}\dfrac{1}{p}$$ is divergent.       I believe that the series $$\sum_{m\text{ is composite}}\dfrac{1}{m}$$ is also divergent.   But I have no idea to attempt for a proof. Any help will be appreciate. Thank you.
34945
I am sorry that my question is not whether $\sum^\infty_{n=2} \frac{1}{\ln{(n!)}}$ convergent or not.  My question is how to compute the sum of series:  $$\sum^\infty_{n=2} \frac{1}{\ln{(n!)}},$$ It is a question in a textbook, and I have worked on it for days, but fail.
118919
If $a_n &gt;0$ for all $n\geq1$, show that the series  $$\sum_{n=1}^{\infty} \frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)}$$  converges.  Can someone check if my solution is correct:  $$\sum_{n=1}^{M}\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-\frac{1}{(1+a_1)(1+a_2)...(1+a_M)} &lt; 1 $$  Since the partial sum of the series is bounded, the series is convergent.
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! :)
69177
Do the following series converge if $a_n&gt;0$ and $\sum_{n=1}^{\infty}a_n$ diverges ?  a.) $\sum_{n=1}^{\infty}{a_n \over 1+ a_n}$  b.)$\sum_{n=1}^{\infty}{a_n\over 1+ a_n ^2}$  c.)$\sum_{n=1}^{\infty}{a_n\over 1+ na_n }$  d.)$\sum_{n=1}^{\infty}{a_n\over 1+ n^2a_n }$  What I thought was using Alembert criteria , but it just not seem to work. I also now if $a_n$ converges so does $a_n^2$. Ive tried everything to no avail, can anyone please help ?
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.
38539
Let $a_n$ be a sequence of positive reals, such that the partial sums $S_n = \sum_{i = 1} ^ n a_i$ diverge to $\infty$. For given $\epsilon &gt; 0$ do we have $$\sum_{n = 1} ^ \infty \frac{a_n}{S_n^{1 + \epsilon}} &lt; \infty?$$  For $\epsilon \ge 1$ we can resolve this quickly by noting $$\frac{a_n}{S_n ^ 2} \le \frac 1 {S_{n - 1}} - \frac 1 {S_n}$$ so for sufficiently large $n$ we can bound $\frac{a_n}{S_n^{1 + \epsilon}}$ by $\frac 1 {S_{n - 1}} - \frac 1 {S_n}$ as well. I'm wondering if this is true for arbitrary $\epsilon &gt; 0$. I know that the series in question diverges for $ \epsilon = 0$, so all that is missing is what happens in $(0, 1)$.
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?
29779
I have to test the convergence of the following series:- $$\sum_{n=1}^\infty\sqrt[3]{n^3+1}-n$$ My approach is as follows :- $$n^3+1&gt;1=\sqrt[3]{n^3+1}&gt;1=\sqrt[3]{n^3+1}-n&gt;1-n$$ Now since$\sum 1-n$ diverges, the series under consideration diverges. Is this right or wrong?
46652
Why does the series $\sum_{n=1}^∞ \ln ({n \over n+1})$ diverges? I'm looking for an answer using the comparison test, I'm just not sure what I can compare it to.   And can I have some tips on what to look at when handling with series that have logarithms in the expression?   Thanks in advance!
97994
I know $4^{n}$ diverges, and $\frac{(n!)^2}{(2n)!}$ converges. I also think that their product diverges since the term of this series is increasing but I don't know how to prove it. I tried ratio test which is inconclusive, and some other tests.
101345
I tried getting it into a closed form but failed. Could someone help me out?  $$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$