Prove that the series $a_n = $ $2^n \over n!$ converges to 0.  I am trying to find a variable that will be bigger than $2^n\over n$ , with no luck.  Any help, besides using binoms? Thanks in advance!!!
1447
Why is     $$\lim_{n \to \infty} \frac{2^n}{n!}=0\text{ ?}$$   Can we generalize it to any exponent $x \in \Bbb R$? This is to say, is     $$\lim_{n \to \infty} \frac{x^n}{n!}=0\text{ ?}$$     This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .
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 ?
43842
Let $\sum a_n$ be a series of non-negative terms and let $$L = \lim_{n\to\infty}n\left(1-\frac{a_{n+1}}{a_n}\right)$$ Prove that the series converges (resp. diverges) if $L &gt; 1$ (resp. $L&lt;1$). I've tried, for example, that when $L&lt;1$, $$n\left(1-\frac{a_{n+1}}{a_n}-L\right)= \frac{n(a_n-a_{n+1}-La_n)}{a_n}\ge\frac{n(a_n-a_{n+1}-a_n)}{a_n}=\frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?  P.S. using Kummer's test doesn't count
31118
If $\sum (a_n)^2$ converges and $\sum (b_n)^2$ converges, does $\sum (a_n)(b_n)$ converge?  Could someone help me to solve this or at least give me a hint?, I have tried using Cauchy's criterion, the Dirichlet test for convergence, etc, but I can´t prove it.Honestly I don´t know where to start. Any help will be appreciated.
44461
I would appreciate your help!  How can we show that if a sequence of positive real numbers $a_n$ converges to $a\in\mathbb{R}$ with $a&gt;0$, then $(a_0 a_1\cdots a_n)^{\frac{1}{n}}$ converges to $a$.
30158
Let $a_n$ be a sequence of nonnegative numbers. Then I want to show that      If $\,\sum_{n=1}^\infty a_n^2\,/ n\,$ converges, then $\frac 1N\!\sum_{n=1}^{N}a_n\,$ converges to zero.    A hint suggests me to write $a_n=\sqrt{n}\cdot \frac{a_n}{\sqrt n}$ and use the Cauchy-Schwarz inequality. However it does not seem enough. I cannot show that $\frac 1N\!\sum_{n=1}^{N}a_n$ converges to zero.  Could anyone please help me to prove it?
10079
Does anyone know how to find the radius of convergence of the series $\sum\ z^{n!}$, where $z$ is a complex number?  I tried to use the definition: $$\frac{1}{R}=\limsup\left|\frac{a_{n+1}}{a_{n}}\right|$$ but I wasn't successful.
95818
$$\sum_{n=0}^\infty n!x^{n^2}$$   Can anyone tell me what will be the radius of convergence for this power series. According to me it is $0$.   If you can please show me how to go about it if I am wrong.
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?
17492
Apparently,   $$ \sum_{n = 0}^\infty \frac{n}{2^n} $$  converges to 2. I'm trying to figure out why. I've tried viewing it as a geometric series, but it's not quite a geometric series since the numerator increases by 1 every term.
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.
95762
As a follow-up to a , I'd like to know if  $$\sum_{n=1}^\infty\frac{1}{2 ^ n}$$  is convergent? If yes, then what does it converge to?  Just as a sidenote - the whole issue started with reading about Zeno's Turtle Paradox which made me curious about the underlying mathematics.
15607
Given that $0\le a_n\lt 1$ the series $\sum_{n=0}^{\infty} (a_n)$ converges. Show that the series $\sum_{n=0}^{\infty} \frac{a_n}{1-a_n}$ converges.  This question is supposed to be solved from first principles (e.g. Comparison Test); however, any approach would be appreciated.  I noted that $\frac{a_n}{1-a_n} = a_n + a_n^2 + a_n^3 + ...$ but I can't seem to finish the proof rigorously.
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.
52454
I'm trying to find out whether sequence $a_n= \frac{n!}{n^n}$ converges or not. My textbook says to compare with $\frac{1}{n}$, and my answer sheet says that $\lim_{x\to \infty} {\frac{n!}{n^n}}\leq \lim_{x\to \infty}\frac{1}{n}$. I don't get this part... Is $\frac{n!}{n^n} \leq \frac{1}{n}$? How did this relationship come about?
4075
I have to prove the  by tomorrow and I am really unconfident about what I did:  $$\sum_{n=1}^\infty a_n\text{ converges } \iff \sum_{n=1}^\infty 2^n a_{2^n}\text{ converges}$$  I did the following:  Let $(b_n)$ be a sequence as follow: $b_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m&lt;2^k$.  It's $a_{n+1}\leq a_n$ and so $0\leq a_{n+p}\leq a_n$ for all $n,p\in\mathbb N$.  So $\sum\limits_{n=1}^\infty b_n$ converges by the majorizing series $\sum\limits_{n=1}^\infty a_n$. And it's $\sum\limits_{n=0}^\infty b_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n+1}}=\sum\limits_{n=1}^\infty 2^{n-1}a_{2^n}$ so $\Rightarrow$ is done.  For $\Leftarrow$ consider $c_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m&lt;2^k$.  It's $|a_n|\leq c_n$ and $\sum\limits_{n=0}^\infty c_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n}}=\sum\limits_{n=1}^\infty 2^{n}a_{2^n}$ and so $\sum\limits_{n=1}^\infty a_n$ converges by the majorizing series $\sum\limits_{n=0}^\infty c_n$.  Is this in form and content correct?
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?