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 $1/\left(1+\frac{1}{n}\right)^n$
2244
I'm trying to find  $$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} .$$   I tried couple of methods: Stolz, Squeeze, D'Alambert  Thanks!  Edit: I can't use Stirling.
95279
I'm self-studying a bit of complex analysis, and I'm attempting to figure out the following.   Suppose $\lim_{n\to\infty}z_n=A$. How can I show that $$ \lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A. $$  Is there a clever way to write the limit to make it more approachable? Thank you.
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.
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.
47236
My friend's trivia league had this math question:  $$\lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right]$$  After computing a few values, one could guess the answer is $e$ = 2.718...But how can we prove that is the limit?  Someone offered up a hand-wavy proof like this:  \begin{align} \lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right] &amp; = \lim_{n \to \infty} \left[\frac{(n+1)(n+1)^{n}}{n^n} - \frac{n \cdot n^{n-1}}{(n-1)^{n-1}} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)\frac{(n+1)^{n}}{n^n} - n\frac{n^{n-1}}{(n-1)^{n-1}} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(\frac{n - 1 + 1}{n-1} \right)^{{n-1}} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(1 + \frac{1}{n-1} \right)^{n-1} \right] \\ &amp;= \lim_{n \to \infty} \left[(n+1)e - n \cdot e \right] \\ &amp;= \lim_{n \to \infty} e \\ &amp;= e \end{align}  The part about substituting $e$ is hand-wavy since technically this is an indeterminate form of $\infty - \infty$. And using $e$ as as upper bound did not lead me to an easy proof either.  Is there a way to rigorously prove the limit? I tried a few approaches: (a) sandwiching the limit--I could prove $e$ was a lower bound, but I could not find a suitable upper bound converging to $e$, (b) using L'Hopital's rule with no luck, (c) using the mean value theorem--but that also got complicated.  So this is a pretty tough problem to ask at trivia! Is there a way to prove this limit formally?  Sources  Trivia question:   Thread on trivia:   Hand-wavy proof:   Idea for mean value theorem:
83831
I dont get it, if I look at $$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$$    It seems like it should be $1$ because $1 \over n$ is almost zero and $1^\infty$ is still $1$
28088
I have the following problem.   I have to find the limit of $b_n = \frac{a_n}{n}$, where $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=l$  My approach:  I express $a_n$ in terms of $b_n$, i.e. $a_n=nb_n$ and $a_{n+1}=(n+1)b_n$  We look at the difference: $a_{n+1}-a_n=(n+1)b_{n+1}-nb_n$  Assuming that $b_n$ converges to a real number m, we see that:  $l=(n+1)m-nm$, from where I conclude that $m=l$.  What I'm left with is proving that $b_n$ is convergent which I'm not sure how to do.  Thanks in advance!
69669
For $a,b,c&gt;0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$  how can I find the limit of sequence above?  Provide me a hint or full solution. thanks ^^
65279
How can I calculate this limit? $$\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n+1}-\sqrt{n}}\cdot(\frac{n+1}{2})!\cdot(\frac{n+1}{2})!}{(n+1)\cdot(\frac{n}{2})!\cdot(\frac{n}{2})!}$$ I don't have idea and I will be happy for help.
20697
I am trying to find the limit of:  $$\lim_{n \to \infty} n(\sqrt[n]{2} - 1)$$  I know it should be very simple but I don't seem to get it. Thanks in advance for any help!
116995
$$\lim_{x\rightarrow +\infty}\sqrt{x}e^{-x}\left(\sum_{k=1}^{\infty}\frac{x^{k}}{k!\sqrt{k}}\right)$$ Any hint will be appreciated.  Note: There is a related question on MathOverflow: . The MO question references this question and it also links to some other posts containing similar expressions.
7484
Let $(a_n)_{n \in \mathbb{N}}$ be a convergent sequence with limit $a \in \mathbb{R}$. Show that the arithmetic mean given by: $$s_n:= \frac{1}{n}\sum_{i=1}^n a_i \tag{A.M.} $$ also converges to $a$.  I have read:  but unfortunately the answers there don't help me much because I don't understand their substitutions and most of it all, why their substitutions seem to work.   What I know from the problem is that since $a_n$ is convergent and $\epsilon &gt;0$ is given, I can say that:   $$\exists N_1 \in \mathbb{N}, \forall n \geq N_1: |a_n-a|&lt;\epsilon_1 $$ I also know that since $a_n$ is convergent, it is bound, so $(a_n) &lt; M, \ \forall n \in \mathbb{N}$   I need to show that: $$\exists N_2 \in \mathbb{N}, \forall n \geq N_2: |s_n-a|&lt; \epsilon_2$$ I started as follows: $$ \left|s_n-a \right| = \left|\frac{1}{n}\sum_{i=1}^na_i -a\right|= \left|\frac{1}{n}\sum_{i=1}^m(a_i-a)+\frac{1}{n}\sum_{i=m+1}^n(a_i-a) \right| \\ \leq \frac{1}{n} \sum_{i=1}^m|(a_i-a)|+\frac{1}{n}\sum_{i=m+1}^n|a_i-a|$$ I believe to understand that the left sum after the $\leq$ is finite, bound and doesn't depend on the values that $n$ takes on. However, I don't understand where all the substitutions come from and make this proof so seemingly easy to complete.   Is there a general idea I can follow to complete such proofs? Because I know that the last step is to show that the given sum is smaller than $\epsilon$. I also know that I should bring the condition $a_n &lt; M$ into place somewhere, but I don't know where.  If I choose an $n \geq N: |a_n-a|&lt; \epsilon'$ what does that tell me about $|a_i-a|$? I know that they are seemingly the same just with a different index.
64999
This was a practice problem I was given.      Show that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$.   My argument is as follows: \begin{align} \lim_{n \rightarrow \infty} \sqrt[n]{a} &amp;= \lim_{n \rightarrow \infty} e^{\ln(\sqrt[n]{a})} \\ &amp;= e^{\lim_{n \rightarrow \infty}(\frac{1}{n}) \ln(a)} \\ &amp;= e^{\lim_{n \rightarrow \infty}(0 \cdot \ln(a))} \\ &amp;= e^0 \\ &amp;= 1.\end{align}
71028
I have to prove that $a_n$ is (strictly) increasing and diverges  $a_n = \sqrt[n]{n!}$ ;  n $\in$  $\ \mathbb{N}$  From sequence I see that $a_n$ increasing to infinitive.  $\sqrt[1]{1!}=1 ,\  \sqrt[2]{2!} \approx 1.41, \  \sqrt[3]{3!}  \approx 1.81 ,...,  \sqrt[n]{n!} $
124275
Let $\{a_n\}$ be a sequence of real numbers such that $$\lim_{n \to \infty}\left|a_n+ 3\left(\frac{n-2}{n}\right)^n \right|^{\frac{1}{n}}=\frac{3}{5}.$$  What is the value of $\lim_{n\to \infty} a_n$?
66113
Let $\ a_n=\sqrt{n+1}-\sqrt{n}$. I have to show that $\lim_{n\to \infty}a_{n}=0$.     How should I start? Do I have to use any theorem?
24947
When $\lim_{n\to\infty} x_n = a$, and $\lim_{n\to\infty} y_n = b$, find the limit, $$\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}.$$ Thank you for your help in advance.