Evaluating limit $ \lim_{n\to \infty} \left(\frac{\frac{a_n}{s_n}}{- \ln (1- \frac{a_n}{s_n})}\right)$ Let $a_n= \sqrt{n},\, n \geq 1,$ and let $s_n= a_1+ a_2....+ a_n.$ Then how to find $$ \lim_{n\to \infty} \left(\frac{\frac{a_n}{s_n}}{- \ln (1- \frac{a_n}{s_n})}\right)$$
 A: All we need to show is that $\frac{a_n}{s_n}\to 0$ as $n\to \infty$.  It is trivial to see that $s_n=\sum_{k=1}^n \sqrt k>n$ so that $\frac{a_n}{s_n}<\frac{1}{\sqrt n}$.  So clearly $\frac{a_n}{s_n}\to 0$.
And inasmuch as $\frac{-x}{1-x}\le \log(1-x)\le -x$ we see that
$$\frac{\frac{a_n}{s_n}}{\frac{\frac{a_n}{s_n}}{1-\frac{a_n}{s_n}}}\le \frac{\frac{a_n}{s_n}}{-\log\left(1-\frac{a_n}{s_n}\right)}\le \frac{\frac{a_n}{s_n}}{\frac{a_n}{s_n}}\tag1$$
Letting $n\to \infty$ in $(1)$ and applying the squeeze theorem we find that 
$$\lim_{n\to\infty}\frac{\frac{a_n}{s_n}}{-\log\left(1-\frac{a_n}{s_n}\right)}=1$$
A: You can use Stolz-Cesaro for the case $\frac{\infty}{\infty}$
$$\lim_{n\to\infty}\frac{a_n}{s_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{s_{n+1}-s_n}=\lim_{n\to\infty}\frac{\sqrt{n+1}-\sqrt{n}}{\displaystyle\sum_{i=1}^{n+1}\sqrt{i}-\displaystyle\sum_{i=1}^n\sqrt{i}}=\lim_{n\to\infty}\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}}\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\\\lim_{n\to\infty}\frac{1}{\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=0$$
Now you can use the standard limit:
$$\lim_{t\to 0}\frac{\ln(1+t)}{t}=0$$ your limit: 
$$\lim_{n\to\infty}\frac{1}{\frac{\ln\left(1-\frac{a_n}{s_n}\right)}{-\frac{a_n}{s_n}}}=1$$
