How to calculate limit of the sequence $e^{-(n^\frac{1}{2})}{(n+1)^{100}}$

Question

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

Answer 1

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, $$\lim_{n\to +\infty}\frac{\ln(n+1)}{\sqrt{n}}=0.$$

Answer 2

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

$$e^{-j}j.$$

Answer 3

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?

Answer 4

If you know expansion of $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ $$\lim_{n \to \infty} e^{-(\sqrt n)}{(n+1)^{100}}=\\ \lim_{n \to \infty} \frac{{(n+1)^{100}}}{e^{\sqrt n}}=\\ \lim_{n \to \infty} \frac{{(n+1)^{100}}}{1+\sqrt{n}+\frac{(\sqrt{n})^2}{2!}+\frac{\sqrt{n}^3}{3!}+...+\frac{\sqrt{n}^{201}}{201!}+....}\to 0\\$$

Answer 5

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 $$\lim_{u \rightarrow \infty} \frac{200! u^0}{e^u}=\lim _{u \rightarrow \infty} 200!~ e^{-\infty}=0.$$

Answer 6

$e^{\sqrt n} \geq \frac {(\sqrt n)^{201}} {(201)!}$. Can you complete the proof from this? [$\frac {(201)! (n+1)^{100}} {n^{201/2}} \to 0$].