How to calculate limit of the sequence $e^{-(n^\frac{1}{2})}{(n+1)^{100}}$ 
Does the sequence $e^{-(n^\frac{1}{2})}{(n+1)^{100}}$ converge? If yes what is the limit?

What I tried: Expanding 
$${(n+1)^{100}}= 1+^{100}C_1n+^{100}C_2{n^2}+^{100}C_3{n^3}+ \dots + ^{100}C_{100}{n^{100}}$$
Multiplying each term by $e^{-(n^\frac{1}{2})}$ and taking limits using L'Hospital Rule we get the limit of the sequence 0, but that is a lengthy and I think it is not a proper approach. [Actually applying L'Hospital rule twice to each term gives the preceding term(I checked it upto $[n^3e^{-(n^\frac{1}{2})}$] and hence ultimately limit will be 0 because lim($e^{-(n^\frac{1}{2})})=0$]
Can anyone please tell me whether this is a correct method to solve the problem and please suggest me a proper method if there is any. Thank you.
The answer is limit of the sequence is 0
 A: Yes, the final limit is zero. Note that as $n\to +\infty$
$$e^{-\sqrt{n}}{(n+1)^{100}}=\exp\left({-\sqrt{n}\underbrace{\left(1-\frac{100\ln(n+1)}{\sqrt{n}}\right)}_{\to 1}}\right)\to0$$
because, for example by using L'Hopital,
$$\lim_{n\to +\infty}\frac{\ln(n+1)}{\sqrt{n}}=0.$$
A: Set $m^2=n$ and you get the product of a negative exponential by a polynomial of degree $200$. The exponential always wins.

Don't be impressed by this exponent,
$$\sqrt[100]{e^{-m}(m^2+1)^{100}}=e^{-m/100}(m^2+1)=10000\,e^{-k}k^2+e^{-k}.$$
As $e^{-k}\to 0$, you can finally reduce to 
$$e^{-j}j.$$
A: Consider
$$f(x)=e^{-(x^\frac{1}{2})}{(x+1)^{100}}$$
and by $x=y^2\to \infty$
$$e^{-(x^\frac{1}{2})}{(x+1)^{100}}=\frac{(y^2+1)^{100}}{e^y}\to 0$$
indeed for any $m€\mathbb N$
$$\frac{y^m}{e^y}\to 0$$
for which you can refer to the related


*

*How to prove that exponential grows faster than polynomial?
A: If you know expansion of $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
$$\lim_{n \to \infty} e^{-(\sqrt n)}{(n+1)^{100}}=\\
\lim_{n \to \infty} \frac{{(n+1)^{100}}}{e^{\sqrt n}}=\\
\lim_{n \to \infty} \frac{{(n+1)^{100}}}{1+\sqrt{n}+\frac{(\sqrt{n})^2}{2!}+\frac{\sqrt{n}^3}{3!}+...+\frac{\sqrt{n}^{201}}{201!}+....}\to 0\\$$
A: Let $n=u^2$, then  $$L=\lim_{u \rightarrow \infty} u^{200} e^{-u} (1+\frac{1}{u^2})^{100} =\lim u^{200} e^{-u}= \lim_{u\rightarrow \infty}\frac{u^{200}}{e^u} \rightarrow \frac{0}{0}.$$ Apply L'Hospital Rule D. w. r. t. $u$ up and down separately 200 times to get $$\lim_{u \rightarrow \infty} \frac{200! u^0}{e^u}=\lim _{u \rightarrow \infty} 200!~ e^{-\infty}=0.$$
A: $e^{\sqrt n} \geq \frac {(\sqrt n)^{201}} {(201)!}$. Can  you complete the proof from this?  [$\frac {(201)! (n+1)^{100}} {n^{201/2}}  \to 0$]. 
