Q- Find the sum of the series -
1+1/3+1/6+1/10+1/15+...........
what should be our approach to solve these kind of questions ?
Thanks,
Ashish
Hi
$\displaystyle S = \frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3}+\cdots+\frac{1}{1+2+\cdots+n}+\cdo ts$
I presume that you know that $\displaystyle 1+2+\cdots+n = \frac{n(n+1)}{2}$
Therefore
$\displaystyle S = \sum_{k=1}^{+\infty}\frac{2}{k(k+1)}$
Spoiler:
Hi again Ashish,
I think you must have guessed that
the general term of this series is $\displaystyle \frac{2}{n(n+1)} $
You need to find
$\displaystyle \sum_{n=1}^{k}\{\frac{2}{n(n+1)} \}$
Now all we are doing here is write every term
$\displaystyle \frac{2}{n(n+1)} $
as
$\displaystyle \frac{2((n+1)-(n))}{n(n+1)} $
$\displaystyle = \frac{2}{n}-\frac{2}{(n+1)} $
this will make many terms cancel like the previous question
-----------------------------------
Final answer is
Spoiler:
use similar approach
Hello AshishFirst, you need to determine a formula for the $\displaystyle n^{th}$ term of the series.
The denominators are $\displaystyle 1, 3, 6, 10, 15, ...$ whose consecutive differences are $\displaystyle 2, 3, 4, 5, ...$
These are simply the sums of the series $\displaystyle 1, 1+2, 1+2+3, 1+2+3+4, ...$ So the $\displaystyle n^{th}$ term is the sum of the series $\displaystyle 1+2+...+n = \tfrac12n(n+1)$
So the original series is:
$\displaystyle 1 + \frac{1}{3}+ \frac{1}{6} + ... + \frac{1}{\tfrac12n(n+1)} + ...$
Using $\displaystyle \Sigma$ notation, then, we have to find $\displaystyle S=\sum_{n=1}^\infty\frac{2}{n(n+1)}$
This you do using Partial Fractions: $\displaystyle \frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1}$
$\displaystyle \Rightarrow S = \sum_{n=1}^\infty\left(\frac{2}{n}-\frac{2}{n+1}\right)$
$\displaystyle = \frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+ ...$
$\displaystyle = 2$
Grandad