# Sequence and series

• Apr 8th 2009, 01:19 AM
a69356
Sequence 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 AM
running-gag
Hi

$S = \frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3}+\cdots+\frac{1}{1+2+\cdots+n}+\cdo ts$

I presume that you know that $1+2+\cdots+n = \frac{n(n+1)}{2}$

Therefore

$S = \sum_{k=1}^{+\infty}\frac{2}{k(k+1)}$

Spoiler:

$S = 2\:\sum_{k=1}^{+\infty} \left(\frac{1}{k}-\frac{1}{k+1}\right) = 2$
• Apr 8th 2009, 01:53 AM
Quote:

Originally Posted by a69356
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 again Ashish(Hi),

I think you must have guessed that

the general term of this series is $\frac{2}{n(n+1)}$

You need to find

$\sum_{n=1}^{k}\{\frac{2}{n(n+1)} \}$

Now all we are doing here is write every term

$\frac{2}{n(n+1)}$

as

$\frac{2((n+1)-(n))}{n(n+1)}$

$= \frac{2}{n}-\frac{2}{(n+1)}$

this will make many terms cancel like the previous question

-----------------------------------

Spoiler:
2n/(n+1)

use similar approach (Happy)
• Apr 8th 2009, 01:55 AM
Sum of series
Hello Ashish
Quote:

Originally Posted by a69356
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

First, you need to determine a formula for the $n^{th}$ term of the series.

The denominators are $1, 3, 6, 10, 15, ...$ whose consecutive differences are $2, 3, 4, 5, ...$

These are simply the sums of the series $1, 1+2, 1+2+3, 1+2+3+4, ...$ So the $n^{th}$ term is the sum of the series $1+2+...+n = \tfrac12n(n+1)$

So the original series is:

$1 + \frac{1}{3}+ \frac{1}{6} + ... + \frac{1}{\tfrac12n(n+1)} + ...$

Using $\Sigma$ notation, then, we have to find $S=\sum_{n=1}^\infty\frac{2}{n(n+1)}$

This you do using Partial Fractions: $\frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1}$

$\Rightarrow S = \sum_{n=1}^\infty\left(\frac{2}{n}-\frac{2}{n+1}\right)$

$= \frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+ ...$

$= 2$