Evaluation of $$\lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}}$$ without using Limit as a sum and stirling Approximation.  $\bf{My\; Try:}$ Using $$\binom{2n}{n} = \sum^{n}_{r=0}\binom{n}{r}^2$$ Using $\bf{Cauchy\; Schwarz}$ Inequality $$\left[\sum^{n}_{r=0}\binom{n}{r}^2\right]\cdot \left[\sum^{n}_{r=0}1\right]\geq \left(\sum^{n}_{r=0}\binom{n}{r}\right)^2 = 2^{2n} = 4^n$$ So $$\frac{4^n}{n+1}&lt;\sum^{n}_{r=0}\binom{n}{r}^2=\binom{2n}{n}$$ But i did not understand how can i calculate upper bound such that i can apply the Squeeze theorem.
19015
$$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$  Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write,   $$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$  $$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$  $$\log\left(2n\right) - \log\left(n\right) = \log(2)$$  Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor.
115753
How can I go about finding the limit of $$\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = \sum_{k = 1}^{\infty} \frac{1}{2^{k+1}}?$$ Could I use the absolute value theorem? I have a feeling it converges to $0$ but I am not sure.
31566
I'd like a hint to show that:  $$\lim \frac{1}{n} \sqrt[n]{(n+1)(n+2) \cdots 2n} = \frac{4}{e} .$$  Thanks.
20698
$$\displaystyle  \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{2} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{2} + 1\right)}}$$  This is from .   If we add one more variable -      $$\displaystyle L(p,q,m,t) = \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{t} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{t} + 1\right)}}$$   Then, the results I got from W|A are really interesting. They seem to follow the pattern which says      $\displaystyle L(p,q,m,t) = \dfrac{1}{t^{|p-q|/m}}$   If it is indeed true, how to prove it? Factorial approximations did not lead me anywhere.
53732
Evaluate $$\lim_{n\to\infty}\frac {qn+1}{qn} \frac {qn+p+1}{qn+p} \dotsm \frac {qn+np+1}{qn+np}$$ where $p\in\mathbb{N}\setminus\{0,1\},q&gt;0.$   Any hints on how to approach this problem in the first place? The answer should be: $\left(\frac {p+q}q\right)^{1/p}$  I have just tried something and got pretty close to the result anyway:  Let $\lim_{n\to\infty} \frac {qn + 1}{qn}\frac {qn+p+1}{qn+p}\dotsm\frac {qn+np+1}{qn+np}=L$ taking the $\ln$ on both sides we get:  $$\lim_{n\to\infty} \sum_{k=0}^n\ln\left(1+\frac1{qn+kp}\right).$$  Here (I'm not sure if I'm wrong here) but if we use the remarkable limit on each of these $\ln$. $\lim_{x\to0}\frac{ln(1+x)}{x}=1$.  $$\lim_{n\to\infty}\sum_{k=0}^n\frac1n\frac1{q+ \frac knp}\xrightarrow[{n\rightarrow\infty}]{}\int_0^1\frac 1{q+xp}dx=\frac 1p\ln\frac {p+q}{q}=\ln L.$$  $\implies \boxed{L=\sqrt[p]{\frac{p+q}q}}.$
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.
39046
I managed to prove it by rewriting the partial products in terms of $n!$ and then applying Stirling's approximation. But I was wondering if there is a more elementary method to show that the limit is zero, without using such heavy machinery.  Also, is it true that in general $$\prod_{n=0}^\infty \frac{an+b}{an+c}=0, \qquad a&gt;0\ \textrm{and}\ c&gt;b$$
52587
Find $$\lim_{n \to \infty} \frac {1}{2^n} \sum_{k=1}^n \frac{1}{\sqrt{k}} \binom{n}{k}.$$  First time I thought about Stirling's approximation but didn't get anything by applying it. I would also think about a Riemann Sum, but no idea how to rewrite...  The answer is $0$.
66365
I am facing difficulty with the following limit. $$ \lim_{n\to\infty}\left(\binom{n}{0}\binom{n}{1}\dots\binom{n}{n}\right)^{\frac{1}{n(n+1)}} $$ I tried to take log both sides but I could not simplify the resulting expression. Please help in this regard. Thanks.
116726
Can anyone help me with this limit problem without L'Hopital rule and Taylor series? $$\lim_{x\rightarrow\ -\infty}\left(-2\frac{\arctan(x)}{\pi}\right)^x$$
111590
I am trying to prove that $\binom{2n}{n} &lt; \frac{4^n}{\sqrt{2n}}$. I tried induction, but with no effect (all I can get to is $(2n+1)(2n+2) &lt; 4\sqrt{n(n+1)}$ which is false)
3540
I don't know what to do with this. Nothing works. I hope somebody can help me to find a decision $$n!\leq {\left(\frac{n+1}{2}\right)}^{n}$$
69025
Evaluate the limit $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$, without using a Riemann sum   $\bf{My\; Try:}$ Using the graph of $\displaystyle f(x) = \frac{1}{\sqrt{x}}\;,$ we get  $$\int_{1}^{n+1}\frac{1}{\sqrt{x}}dx &lt;\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)&lt;1+\int_{1}^{n}\frac{1}{\sqrt{x}}dx$$  So we get $$2\sqrt{n+1}-2&lt;\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)&lt;2\sqrt{n}-1$$  So $$\lim_{n\rightarrow \infty}\frac{2\sqrt{n+1}-2}{\sqrt{n}}=2&lt;\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)&lt;\lim_{n\rightarrow \infty}\frac{2\sqrt{n}-1}{\sqrt{n}} = 2$$  So using the Sandwich Theorem, we get $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)=2$  My question is can we solve it by using any other method? If yes then please explain it here.
94341
How to show that $\sum\limits_{n = 0}^{\infty} \frac {1}{n!} = e$ where $e = \lim\limits_{n\to\infty} \left({1 + \frac 1 n}\right)^n$?   I'm guessing  this can be done using the Squeeze Theorem by applying the AM-GM inequality. But I can only get the lower bound.   If $S_n$ is the $n$th partial sum of our series then,   $$ \left({1 + \frac 1 n}\right)^n = 1 + n\cdot\frac{1}{n} + \frac{n(n - 1)}{2}\cdot \frac{1}{n^2} + \cdots + \frac{n(n - 1)\ldots(n - (n -1))}{n!} \cdot \frac{1}{n^n} $$  $$ \le 1 + \frac{1}{2!} + \cdots + \frac{1}{n!} =S_n $$  How can I show that $S_n \le $ a subsequence of $\left({1 + \frac 1 n}\right)^n$ or any sequence that converges to $e$?
110771
I am trying to show that $\lim_{n\to \infty}{\frac{2n\choose n}{4^n}}=0$. I found that using stirling's approximation, I can get: $$ \lim_{n\to \infty}{\frac{2n\choose n}{4^n}}= \lim_{n\to \infty}{\frac{(2n)!}{n!^2*2^{2n}}}= \lim_{n\to \infty}{\frac{(\frac{2n}{e})^{2n}\sqrt{2\pi (2n)}}{((\frac{n}{e})^n\sqrt{2\pi n})^2*2^{2n}}} =\lim_{n\to \infty}{\frac{2\sqrt{\pi n}}{2\pi n}} =0 $$  But this seems inelegant where there should be a more elegant, combinatorial proof. Is there an easier way?
90125
Find $\lim _{ n\rightarrow \infty  }{ { \left( { 3 }^{ n }+1 \right)  }^{ \frac { 1 }{ n }  } } $ using the squeeze theorem  I have come across ways to do this but none mention the squeeze (or sandwich) theorem. I know I need to find $2$ functions which squeeze the given function but can only think of using $(3^n)^{1/n}$ i.e. $3$ as the $\le $ function
3571
Evaluate $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+3)}$.   $\bf{My\; Try::}$ Although we can solve it by converting into definite Integration.  But I want to solve it without Using Integration.  So $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+3)} = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{(n-1)(n-2).......(n-k+1)}{k!\cdot n^{k}\cdot (k+3)}$  Now How can i solve after that, Help required, Thanks