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I wonder if the sequence{$S_n$} where $S_n=\sum_{k=1}^{n}{1\over {\sum_{i=1}^{k} \frac{(i)(i+1)}{2}}}$is bounded and has a limit


Calculate $1+{1\over {1+3}}+ {1 \over {1+3+6}}+...+{1\over {1+3+6+...+5050}}$

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Is this homework? If so, you should take it as such. Even if it's not, you should give some idea as to what you've tried or where you're stuck, and some users do not like it when you give commands (e.g. "Calculate _____"). – Tyler Mar 21 '12 at 3:58
up vote 2 down vote accepted

First note that $$\begin{eqnarray} \sum_{i=1}^{k} \frac{(i)(i+1)}{2}&=& \frac{1}{2}\left(\sum_{i=1}^{k} i^2 + \sum_{i=1}^{k} i\right)\\ &=&\frac{1}{2}\left(\frac{k(k+1)(2k+1)}{6}+\frac{k(k+1)}{2}\right)\\ &=&\frac{ k(k+1)(k+2)}{6}\end{eqnarray}$$ so we have $$S_n=6\sum\limits_{k=1}^n \frac{1}{ k(k+1)(k+2)}\leq 6\sum\limits_{k=1}^n \frac{1}{k^2}\leq \pi^2$$ thanks to Euler's celebrated result that $\sum\limits_{k=1}^n \frac{1}{k^2}\to \frac{\pi^2}{6}$. Thus the sequence $(S_n)$ is bounded, and since it is clearly increasing it has some limit.

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First, $$\sum_{i=1}^k \frac{i(i+1)}2 = \frac12 \left(\sum i + \sum i^2 \right) = \frac{k(k+1)(k+2)}{6}.$$

We are looking at summing up the reciprocal of this as $k$ varies. The reciprocal can be decomposed:

$$\frac6{k(k+1)(k+2)} = \frac{3}{k} + \frac{-6}{k+1} + \frac{3}{k+2}.$$

The sum

$$S_n = \sum_{k=1}^n \left( \frac{3}{k} + \frac{-6}{k+1} + \frac{3}{k+2}\right)$$

telescopes, leaving only

$$\frac31 + \frac{-6+3}2 + \frac{3+(-6)}{n+1} + \frac{3}{n+2} = \frac32 - \frac{3}{n+1} + \frac{3}{n+2}$$

for $n > 1$. This clearly has an upper bound and limit of $3/2$ as $n$ goes to infinity. The requested sum, $S_{100}$, is

$$\frac32 - \frac{3}{101} + \frac{3}{102} = \frac{2575}{1717} \approx 1.4997.$$

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Simpler and less accurate:

The first sum has denominators that are greater than a constant times $n^3$, and the second sum has denominators that are greater than a constant times $n^2$.

In both cases, since all terms are positive, and $\sum 1/n^2$ converges, the sum is bounded, converges, and must have a limit.

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