Convergence of $a_n=(n+1)^{100}e^{-\sqrt{n}}$ for $n\geq 1$ Question is to check for  Convergence of :
$a_n=(n+1)^{100}e^{-\sqrt{n}}$ for $n\geq 1$
what i could do is :
 $$\frac{a_{n+1}}{a_n}=\frac{(n+2)^{100}e^{-\sqrt{n+1}}}{(n+1)^{100}e^{-\sqrt{n}}}=\frac{(1+\frac{1}{n+1})^{100}}{e^{\sqrt{n+1}-\sqrt{n}}}$$
Numerator goes to $1$ and denominator goes to infinity so, limit might go to $0$
I am sure that i should not check limit of numerator and denominator differently but i could not use any other way to proceed.
Please help me to clear this.
Thank you :)
 A: Hint: if $t = \sqrt{n}$, $a_n = (t^2 + 1)^{100} e^{-t} < $.  Exponentials grow faster than polynomials...
A: For n is big enough $(n+1)^{100}n^2\leq e^{\sqrt{n}}$
So $a_n=\frac{(n+1)^{100}}{e^{\sqrt{n}}}\leq \frac{1}{n^2}$ for large n
So by comparing theorem, we can judge this sequence is convergence.
A: We will use the fairly well-known inequality, for all $x$,
$$
1+x\le e^x\tag{1}
$$
Assume $x\gt-1$. Multiplying both sides by $e^{-x}$, substituting $x=t/201$, raising both sides to the $201^\text{st}$ power, and dividing by $1+t/201$ yields
$$
(1+t/201)^{200}e^{-t}\le\frac1{1+t/201}\tag{2}
$$
Thus, for $t\ge0$,
$$
\begin{align}
\left(1+t^2\right)^{100}e^{-t}
&\le40401^{100}(1+t^2/40401)^{100}e^{-t}\\[3pt]
&\le40401^{100}(1+t/201)^{200}e^{-t}\\
&\le\frac{40401^{100}}{1+t/201}\tag{3}
\end{align}
$$
Inequality $(3)$ tells us that
$$
(1+n)^{100}e^{-\sqrt{n}}\le\frac{40401^{100}}{1+\sqrt{n}/201}\tag{4}
$$
Therefore,
$$
\lim_{n\to\infty}(1+n)^{100}e^{-\sqrt{n}}=0\tag{5}
$$
A: For large values of "n", Exp[ Sqrt[n+1] - Sqrt[n] ] has a Taylor expansion which write
1 + 1 / (2 Sqrt[n]) + 1 / (8 n) + ...
So, the denominator goes to 1 and not to zero.
Concerning the overall ratio a(n+1) / a(n), for large values of "n", it varies as
1 - 1 / (2 Sqrt[n]) + 801 / (8 n) + ...  
