I have just approached the following series  $$\sum_{n=1}^\infty \sin [\pi(\sqrt5+2)^n]$$ And I already have a question. The $\lim_{n \to \infty}\pi(\sqrt5+2)^n=+\infty$. And the $\lim_{n\to \infty} \sin[\pi(\sqrt5+2)^n]$ does not exist.  Can $\sum a_n$ converge if $\lim_{n\to \infty}a_n$ does not exist? I guess the answer is yes because I know that $\sum a_n$ cannot converge if $\lim_{n\to \infty}a_n\neq 0$. Am I right?    The second part of my question concerns the strategy to use to test convergence of such a series. I can test convergence of series with standard tests. But it seems I need to develop a more sophisticated approach in order to successfully deal with series like this. For example, here, should I use some sort of expansion? If yes, what kind of expansion?  Thank you for your help.
121486
Per the title, does $\displaystyle\sum_{n=1}^{\infty} \sin(\pi(2+\sqrt{3})^n)$ converge? Converge absolutely?   I'm stuck on this question, not sure how to approach it.  Thank you!
88110
I am given the problem: Let $a_n&gt;0$, prove $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \sin(a_n)$ converges.  We have this problem in a homework, and I don't believe it can be true. The statement being biconditional implies that when $\sum_{n=1}^{\infty} \sin(a_n)$ converges, so too does $\sum_{n=1}^\infty a_n$, but if we take $a_n=\pi n$ or even $a_n=\pi$, then $\sum_{n=1}^\infty \sin(a_n)$ converges while $\sum_{n=1}^\infty a_n$ diverges. Is my thinking incorrect?
34983
$$\sum_{n=1}^\infty {{4^n n!n!}\over{(2n)!}}$$  I tried the ratio test but got that the limit is equal to 1, this tells me nothing of whether the series diverges or converges. if I didn't make any errors when doing the ratio test,  it may diverge, but I need help proving that. Is there any other test I could try.
1158
Does $\sum_{k=1}^\infty |\sin(ak)/k|$ converge for all $0&lt;a&lt;\pi$? I do not think so since for $a=\pi /2$: $$\sum_{k=1}^\infty\left\vert\frac{\sin(ak)}{k}\right\vert=\sum_{k=0}^\infty (2k+1)^{-1}=\infty.$$  But is this true for all $a$?
103129
I have a function defined as: $$S(a)=\sum_{n=1}^\infty\frac{\sin(n)}{n^a}$$ My question is for what values of $a$ is this convergent, and how can I evaulate this?  For starters, I know that $S(a)$ is definately convergent  for a $a\ge2$ since: $$\sum_{n=1}^\infty\frac{\sin(n)}{n^a}\le\sum_{n=1}^\infty\frac{1}{n^a}$$ So we can use $\zeta(a)$ as the upper limit for the series, which makes me think I can extend this convergence to $a&gt;1$. However, I have no idea how I would evaluate this. Thanks!
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.
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.
75893
This is the sum   $$\sum_{n=1}^{\infty}\frac{1}{{5^n}^!}$$  My first attempt was to assume that the series does converge to a rational number $a/b$. But the $n!$ bothered me and I failed in my proof.  How would you try to prove this series?
43031
I am trying to conclude about the convergence of $$\sum_{n=1}^{\infty}{\sqrt[n]{n}-1 \over n}$$ If I take ${\sqrt[n]{{\sqrt[n]{n}-1 \over n}}}= {\sqrt[n]{\sqrt[n]{n}-1} \over \sqrt[n]{n}}$ this looks like it converges to $0$ but I am suspicious about the numerator. At the same time, other techniques I know like the ratio test, $n(\sqrt[n]{n}-1)$ and comparisons don't yield meaningful results. Any hints?
44542
How can one prove that the series $\displaystyle \sum\limits_{n=1}^\infty \frac{1}{(n!)^2}$ converges to an irrational number? There's no need to use Taylor expansion, integrals or any advanced/professional techniques. It can be proved using only basic techniques. My first attempt was to assume that the series does converge to a rational number $ \frac{p}{q}$ and then it follows by definition that for every positive epsilon there exists an integer k such that $ {\sum_{n=1}^{k} \frac{1}{(n!)^2} -\frac{p}{q}}&lt;=\epsilon$ I tried to get a contradiction but I failed.
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)...
123584
Sorry for the oversimplified question, but does the series $\sum_{n=1}^{\infty} \frac{1}{n+1}$ converge? The ratio test of it gives the result of "1". Thanks a lot.
61887
I know that the series $\sum_{k=1}^\infty \frac{\sin k}{k}$ converges (to $\frac{\pi - 1}{2}$), though by crazy stuff with Dirichlet Kernels or by reverse-engineering $\frac{\pi - x}{2} = \sum \frac{\sin (kx)}{k}$ using Fourier Series.  My question instead: Is $\sum_{k=1}^\infty \frac{\sin k}{k}$ absolutely convergent?  My gut instinct is no: It behaves not-much-better than $\sum \frac1k$, but that's certainly not a rigorous proof.
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 ?
96253
Just out of curiosity, I was wondering if anybody knows any methods (other than the integral test) of proving the infinite series where the nth term is given by $\frac{1}{n^2}$ converges.
40706
Let $a_n$ be a sequence. $\lim_{n\to\infty} a_n =0 $ then the series: $\sum_{n=1}^\infty a_n, \ \sum_{n=1}^\infty (a_n+a_{n+1}) $ converge and diverge together.  It doesn't seem right, there are no convergence tests that fit this question yet every counterexample I try does not seem to work. Is this how I should see the sum? $\sum_{n=1}^\infty (a_n+a_{n+1})=2\sum_{n=1}^\infty a_n +a_{n+1}$ If so then there's no real difference between the two series since you multiply it by a constant and add just one more value.
109643
Does $\sum\frac{\sin n}{n}$ converge?  I have tried the comparison test, root test and ratio test but still can't prove it is convergent or divergent.