Find $\lim\limits_{x\to\infty}\left(\frac{a_1}{a_0S_1}+\frac{a_2}{S_1S_2}+...+\frac{a_n}{S_{n-1}S_n}\right)$ 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)?

  • 2
    $\begingroup$ May be your Question is $$\lim\limits_{n\to\infty}\left(\frac{a_1}{S_0S_1}+\frac{a_2}{S_1S_2}+...+\frac{a_n}{S_{n-1}S_n}\right)$$ $\endgroup$
    – juantheron
    Sep 18 '15 at 17:35
  • 5
    $\begingroup$ Try to exploit $a_{n}=S_n-S_{n-1}$ to get a telescopic sum. $\endgroup$ Sep 18 '15 at 17:37
  • $\begingroup$ No. It seems that would simplify this. $\endgroup$
    – user300045
    Sep 18 '15 at 17:38
  • $\begingroup$ @Jack D'Aurizio I can't get to the telescopic sum. $\frac{a_1}{a_0S_1}+\sum\limits_{k=2}^{n}\left(\frac{S_k}{S_{k-1}S_k}-\frac{S_{k-1}}{S_{k-1}S_k} \right)$. $\endgroup$
    – user300045
    Sep 18 '15 at 18:38

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}

  • $\begingroup$ I have to say the first lines are quite irrelevant, the telescopic sum just follows from $a_k= S_k-S_{k-1}$, no matter if $S_k$ has some nice closed form or not, we just need $S_n\to +\infty$ as $n\to +\infty$. $\endgroup$ Sep 18 '15 at 20:24

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 $1$.


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