Limit of the infinite sequence involving $a_n, s_n$ 
Let $a_n=\sqrt{n}$ and $s_n=a_1+a_2+\ldots+a_n$. Find the limit : $$\lim_{n\to +\infty}\left[\frac{\frac{a_n}{s_n}}{-\ln(1-\frac{a_n}{s_n})}\right]$$

I am hopelessly confused with this problem. Is L'Hopital or Cesaro-Stolz of any use here? Any hints. Thanks beforehand.
 A: We have that by Stolz-Cesàro:
$$
\lim_{n\to +\infty} \frac{a_n}{s_n}=\lim_{n\to +\infty} \frac{a_{n+1}-a_{n}}{s_{n+1}-s_{n}}=\lim_{n\to +\infty} \frac{a_{n+1}-a_{n}}{a_{n+1}}=\lim_{n\to +\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}}=0
$$
Therefore:
$$
\lim_{n\to +\infty}\frac{\frac{a_n}{s_n}}{-\ln\left(1-\frac{a_n}{s_n}\right)}=\lim_{x\to 0}\frac{x}{-\ln\left(1-x\right)}=1
$$
A: Observe that $x_n=a_n/s_n$ converges to zero. Then, the limit becomes trivially solvable by Taylor expansion of the logarithm.
Estimate:
$$s_n=\sum_{k=0}^n \sqrt{n}\approx \int_{0}^n \sqrt{n}dn=\frac{2}{3}n^{3/2}$$
(we are only interested in the power scaling - we need to see that $3/2>1/2$, so we don't need to talk about the error, but you can estimate the error from Euler-Maclaurin formula, which then proves that all the extra terms are negligible in the limit.
So you have
$$x_n=\frac{a_n}{s_n}\asymp\frac{1}{n}\underset{n\to \infty}{\to} 0 $$
By the way, you don't have to use integrals to show $a_n/s_n\to 0$, I just find it elegant. You can use induction or upper bounds or anything similar.
A: Observe that $$\lim_{t\to 0} \frac{t}{-\ln (1-t)} =\lim_{t\to 0} \frac{1}{(1-t)^{-1}} =\lim_{t\to 0} (1-t)=1$$
So by Heine definition of limit we have that for any sequence $x_n $ tending to $0$ the following equality holds $$\lim_{n\to \infty} \frac{x_n}{-\ln (1-x_n)}=1$$
