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

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- Apr 8th 2009, 01:19 AMa69356Sequence and series
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 - Apr 8th 2009, 01:50 AMrunning-gag
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__: - Apr 8th 2009, 01:53 AMADARSH
Hi again Ashish(Hi),

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 (Happy) - Apr 8th 2009, 01:55 AMGrandadSum of series
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

- Apr 8th 2009, 02:52 AMa69356
Thanks a lot to all of you for quick reply.... (Clapping)