nth partial sum

**krzyrice** · May 24th 2010, 08:55 PM

Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

1)

I have no idea how to do this one. I missed it on the test and it might show up on the final

**chisigma** · May 24th 2010, 09:04 PM

The 'key' is the identity...

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

Kind regards

$\chi$ $\sigma$

**krzyrice** · May 24th 2010, 09:19 PM

i'm still not getting it

**harish21** · May 24th 2010, 09:36 PM

Originally Posted by krzyrice

Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

1)

I have no idea how to do this one. I missed it on the test and it might show up on the final

$\frac{7}{(n+1)(n+2)} = \frac{7}{n+1}-\frac{7}{n+2}$

Now,

$\sum_{n=1}^k \frac{7}{n+1}-\frac{7}{n+2} = \left( \frac{7}{2}-\frac{7}{3} \right) + \left( \frac{7}{3}-\frac{7}{4} \right)+....+ \left( \frac{7}{k+1}-\frac{7}{k+2}\right)$

$= \left( \frac{7}{2} -\frac{7}{k+2}\right)$

find the limit as k tends to infinity.

**krzyrice** · May 24th 2010, 10:56 PM

thanks makes sense now

**Prove It** · May 24th 2010, 11:05 PM

Originally Posted by krzyrice

Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

1)

I have no idea how to do this one. I missed it on the test and it might show up on the final

Surely this is

$7\sum_{n = 2}^{\infty}\frac{1}{n(n + 1)}$ .

Using partial fractions:

$\frac{A}{n} + \frac{B}{n + 1} = \frac{1}{n(n + 1)}$

$\frac{A(n + 1) + Bn}{n(n + 1)} = \frac{1}{n(n + 1)}$

$A(n + 1) + Bn = 1$

$(A + B)n + A= 0n + 1$ .

So $A = 1$ and $B = -1$ .

Therefore we can write the sum as

$7\sum_{n = 2}^{\infty}\frac{1}{n(n + 1)} = 7\sum_{n = 2}^{\infty}\left(\frac{1}{n} - \frac{1}{n + 1}\right)$

This is a telescoping series, so all terms except the first and last will cancel.

So the $n^{\textrm{th}}$ partial sum is

$7\left(\frac{1}{2} - \frac{1}{n + 1}\right)$ .

Therefore, the sum of the series is

$\lim_{n \to \infty}7\left(\frac{1}{2} - \frac{1}{n + 1}\right)$

$= 7\left(\frac{1}{2} - 0\right)$

$= \frac{7}{2}$ .

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