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
Also
Calculate $1+{1\over {1+3}}+ {1 \over {1+3+6}}+...+{1\over {1+3+6+...+5050}}$
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
Also
Calculate $1+{1\over {1+3}}+ {1 \over {1+3+6}}+...+{1\over {1+3+6+...+5050}}$
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.
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
$\frac32 - \frac{3}{101} + \frac{3}{102} = \frac{2575}{1717} \approx 1.4997.$
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.