Why is $\lim\limits_{n\to\infty}(n!)^{1/n}=\infty$   It is more or less clear that the sequence is increasing by ratio test $\frac{((n+1)!)^{1/(n+1)}}{(n!)^{1/n}}=\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n+1]{(n)!}}\cdot\frac{\sqrt[n+1]{(n)!}}{\sqrt[n]{(n)!}}=\sqrt[n+1]{n+1}\cdot(n!)^{\frac{1}{n(n+1)}}$  but left multiplicand tends to $1$, what about the right one ?   Is there convergence/divergence tests for sequences, I only found some for series ?
108053
Find $\lim_{n\to\infty} ((n!)^{1/n})$. The question seemed rather simple at first, and then I realized I was not sure how to properly deal with this at all. My attempt: take the logarithm,  $$\lim_{n\to\infty} \ln((n!)^{1/n}) =  \lim_{n\to\infty} (1/n)\ln(n!) =  \lim_{n\to\infty} (\ln(n!)/n)$$ Applying L'hopital's rule:  $$\lim_{n\to\infty} [n! (-\gamma + \sum(1/k))]/n! = \lim_{n\to\infty} (-\gamma + \sum(1/k))=  \lim_{n\to\infty} (-(\lim(\sum(1/k) - \ln(n)) + \sum(1/k))  = \lim_{n\to\infty} (\ln(n) + \sum(1/k)-\sum(1/k)  = \lim_{n\to\infty} (\ln(n))$$ I proceeded to expand the $\ln(n)$ out into Maclaurin form $$\lim_{n\to\infty} (n + (n^2/2)+...) = \infty$$ Since I $\ln$'ed in the beginning, I proceeded to e the infinity  $$= e^\infty  = \infty$$  So am I write in how I approached this or am I just not on the right track? I know it diverges, I was just wanted to try my best to explicitly show it.
65629
I came across this strange limit whilst showing convergence of a series: $$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$$ How can I calculate this limit?
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.
17721
How to test whether $\sum_{n=1}^{\infty }\frac{1}{n^{1+\frac{1}{n}}}$ diverges or converges? Can I apply ratio test? Thanks in advance.
21003
Prove or disprove: if $\sum\limits_{n=0}^\infty a_n$  converges and $\lim\limits_{n→\infty}⁡b_n =0$, then $\sum\limits_{n=0}^\infty a_n  b_n$ converges.  I tried this example $$a_n= b_n = \frac {(-1)^n} {\sqrt {n+1}},$$ this will show that $$\sum_{n=0}^\infty a_n  b_n = \sum_{n=0}^\infty \frac {1} { {n+1}},$$ which is the Harmonic series, diverges.   However, I tried every convergent test but I couldn't show that $$\sum_{n=0}^\infty \frac {(-1)^n} {\sqrt {n+1}}$$ converges.
12006
Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$  So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$
7590
I am trying to check whether or not the sequence $$a_{n} =\left\{\frac{n^n}{n!}\right\}_{n=1}^{\infty}$$ is bounded, convergent and ultimately monotonic (there exists an $N$ such that for all $n\geq N$ the sequence is monotonically increasing or decreasing). However, I'm having a lot of trouble finding a solution that sufficiently satisfies me.  My best argument so far is as follows,  $$a_{n} = \frac{n\cdot n\cdot n\cdot \ldots\cdot n}{n(n-1)(n-2)(n-3)\dots(2)(1)} = \frac{n}{n}\cdot \frac{n}{n-1}\cdot  \ldots \cdot \frac{n}2\cdot n$$  so $\lim a_{n}\rightarrow \infty$ since $n&lt;a_{n}$ for all $n&gt;1$. Since the sequence is divergent, it follows that the function must be ultimately monotonic.  This feels a little dubious to me, I feel like I can form a much better argument than that, or at the very least a more elegant one. I've tried to assume $\{a_{n}\}$ approaches some limit $L$ so there exists some $N$ such that  $|a_{n} - L| &lt; \epsilon$ whenever $n&gt;N$ and derive a contradiction, but this approach got me nowhere.  Finally, I've also tried to use the fact that $\frac{a_{n+1}}{a_n}\rightarrow e$ to help me, but I couldn't find an argument where that fact would be useful.
50422
I have $\displaystyle\sum_{n=1}^{\infty}{\frac{(2n)!}{(4^n)(n!)^2(n^2)}}$ and need to show whether it diverges or converges.   I attempted to use the ratio test, but derived that the limit of $\dfrac{a_{n+1}}{a_n}=1$ and hence the test is inconclusive.   So I now must attempt to use the comparison test, but I am struggling to find bounds to compare it to to show either divergence or convergence.
106235
Evaluate the limit : $$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)$$  I can use the sandwich principle, certain convergence criteria, Cesaro mean theorem, limit arithmetic.. things around this area.  Any help would be greatly appreciated, thanks!  Sorry for not elaborating more at the beginning, rookie first-post mistake I suppose. :)
68710
I am trying to prove that the sequence $$a_n = (n!)^{\frac2n}$$ tends to infinity as $ n \to \infty $.  I've tried different methods but I haven't really got anywhere. Any solutions/hints?
43522
I think should be something very simple... $$\lim_{n\to\infty}nq^{n}$$ wher $|q|&lt;1$ i tried to use binomial theorem, but no success... I know that $q^{n}$ is convergent to $0$, but can't deal with this $n$
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?
30401
I want to prove the following theorem     Let $\{a_{n}\}$ be a convergent sequence. Prove that $\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}$   Do I need to show that it is monotone increasing and then use the relation  $$0\leq \lim_{n\to \infty}a_{n+1}- \lim_{n\to \infty}a_{n}\leq 0 \;?$$  Please, I need help on this! Various proofs are welcome! Thank you very much for your time!
71352
I am asked to determine if a series converges or not:  $$\displaystyle\sum\limits_{n=1}^{\infty} \frac{(2^n)n!}{(n^n)}$$  So I'm using the $n$th root test and came up with $\lim_{n \to {\infty}}\frac{2}{n}\times(\sqrt[n]{n!})$ I know that the limit of $\frac{2}{n}$ goes to $0$ when $n$ goes to infinity but what about the $(\sqrt[n]{n!})$?
48461
How to prove the following limit?   $$\lim_{n \to \infty} (1+1/n)^n = e$$    I can only observe that the limit should be a very large number!   Thanks.
82532
I have a sequence defined by $$ a_1=1,\quad a_2=1,\quad a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\text{ for } n\ge2. $$ Now, if $\lim_{n\to \infty}a_n=g$ then $\lim_{n\to \infty}a_n=\lim_{n\to \infty}\Bigl(\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\Bigr)=\frac{2}{g}$ so $g=\sqrt{2}$ or $g=-\sqrt{2}$ but $a_n&gt;0$ so $g=\sqrt{2}$.  Now, how do I prove that it has an actual  limit? Also, it can be proven that $1\le a_n\le2$, and it's not monotonic because $a_4 \gt a_5 \lt a_6$. Also, it's not monotonic after some $N\in\mathbb N$
30834
I am trying to evaluate the $\lim(\sqrt[n]{n!})$ using 2 theorems (2 proofs)   Theorem 1: Let $\{c_n\}$ be any sequence in $\mathbb{R}^+$. Then, $\displaystyle \underline{\lim}\frac{c_{n+1}}{c_n}\leq \underline{\lim}\sqrt[n]{c_n}$ and $\displaystyle \overline{\lim}\sqrt[n]{c_n}\leq \overline{\lim}\frac{c_{n+1}}{c_n}$.    so with 1. I have $\frac{(n+1)!}{n!}$ =$n+1$  which is $\overline{\lim}=\infty$   and 2 with $\sqrt[n]{n!}\geq\sqrt[n]{(n/2)^{n/2}}$=$\sqrt{\frac{n}{2}}$ which is $\overline{\lim}=\infty$  is it valid?  P.S I was not using theorem 1 right