Find $\lim\limits_{n\to\infty}\left(\frac{a_1}{a_0S_1}+\frac{a_2}{S_1S_2}+...+\frac{a_n}{S_{n-1}S_n}\right)$ Problem: Find $\lim\limits_{n\to\infty}\left(\frac{a_1}{a_0S_1}+\frac{a_2}{S_1S_2}+...+\frac{a_n}{S_{n-1}S_n}\right)$ where $n=0,1,2,...$ and $a_n=2015^n,S_n=\sum\limits_{k=0}^{n}a_k$
$S_n$ can be written as the geometric sum $S_n=\frac{2015^{n+1}-1}{2014}$.
Applying the values for $a_k$ and $S_k$ can't give a closed form in the limit.
How to transform sequence in the limit so it gives closed form (if possible)?
 A: Since $$S_{n}=\sum_{k=0}^n2015^k=\sum_{k=0}^{n-1}2015^k+2015^n=S_{n-1}+2015^n\qquad n\ge 1$$
it follows
\begin{align}
\frac{a_n}{S_{n-1}S_n}&=\frac{2015^n}{S_{n-1}(S_{n-1}+2015^n)}\\
&=\frac{S_{n-1}+2015^n-S_{n-1}}{S_{n-1}(S_{n-1}+2015^n)}\\
&=\frac{1}{S_{n-1}}-\frac{1}{S_{n-1}+2015^n}\\
&=\frac{1}{S_{n-1}}-\frac{1}{S_n}
\end{align}
Thus
$$\sum_{k=0}^n\frac{a_k}{S_{k-1}S_k}=\underbrace{\left(\frac{1}{S_0}-\frac{1}{S_1}\right)+\left(\frac{1}{S_1}-\frac{1}{S_2}\right)+\ldots+\left(\frac{1}{S_{n-1}}-\frac{1}{S_n}\right)}_{n\text{ terms}}=\frac{1}{S_0}-\frac{1}{S_n}$$
\begin{align}
\lim_{n\to\infty}\sum_{k=0}^n\frac{a_k}{S_{k-1}S_k}&=\lim_{n\to\infty}\left(\frac{1}{S_0}-\frac{1}{S_n}\right)
\end{align}
A: Since $a_k = S_k - S_{k - 1}$ we easily deduce that $$\frac {a_k} {S_{k - 1} S_k} = \frac {1} {S_{k - 1}} - \frac {1} {S_k}.$$ Then, $$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac {a_k} {S_{k - 1} S_k} = \lim_{n \to \infty} \left(\frac {1} {S_0} - \frac {1} {S_n} \right) = \frac {1} {S_0},$$
which I take to be $1$.
