# Thread: nth partial sum of series

1. ## nth partial sum of series

Hi. I'm trying to understand an example in a chapter on Series and Convergence. The example states rather casually that:

The nth partial sum of the series

$\sum^{\infty}_{n=1}{(\frac{1}{n}-\frac{1}{n+1})} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ...$

is given by

$S_{n} = 1 - \frac{1}{n+1}$

But I'm not understanding how we we came to this conclusion. Can somebody help me understand this?

2. ## Re: nth partial sum of series

Originally Posted by infraRed
Hi. I'm trying to understand an example in a chapter on Series and Convergence. The example states rather casually that:

The nth partial sum of the series

$\sum^{\infty}_{n=1}{(\frac{1}{n}-\frac{1}{n+1})} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ...$

is given by

$S_{n} = 1 - \frac{1}{n+1}$

But I'm not understanding how we we came to this conclusion. Can somebody help me understand this?
${S_N} = \sum\limits_{n = 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right]}\\ = \left[ {\frac{1}{1} - \frac{1}{2}} \right] + \left[ {\frac{1}{2} - \frac{1}{3}} \right] + \cdots + \left[ {\frac{1}{N} - \frac{1}{{N + 1}}} \right] \\= 1 - \frac{1}{{N + 1}}$

3. ## Re: nth partial sum of series

Note that except the first and the last term of the partial sum, every term cancels the next term in the serie.

4. ## Re: nth partial sum of series

Originally Posted by Plato
${S_N} = \sum\limits_{n = 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right]}\\ = \left[ {\frac{1}{1} - \frac{1}{2}} \right] + \left[ {\frac{1}{2} - \frac{1}{3}} \right] + \cdots + \left[ {\frac{1}{N} - \frac{1}{{N + 1}}} \right] \\= 1 - \frac{1}{{N + 1}}$
Could you break this down some more? I don't see how 1/1 + 1/2 + ... + 1/N = 1, or how -1/2 - 1/3 - ... - 1/(N+1) = -1/(N+1). Presumably that is not what is happening, but whatever is happening I don't quite see it.

5. ## Re: nth partial sum of series

Originally Posted by infraRed
Could you break this down some more? I don't see how 1/1 + 1/2 + ... + 1/N = 1, or how -1/2 - 1/3 - ... - 1/(N+1) = -1/(N+1). Presumably that is not what is happening, but whatever is happening I don't quite see it.
$\left[ {\frac{1}{1} - \frac{1}{2}} \right] + \left[ {\frac{1}{2} - \frac{1}{3}} \right] + \cdots + \left[ {\frac{1}{N} - \frac{1}{{N + 1}}} \right]\\ = 1 + \left( { - \frac{1}{2} + \frac{1}{2}} \right) + \left( { - \frac{1}{3} + \frac{1}{3}} \right) \cdots \left( { - \frac{1}{N} + \frac{1}{N}} \right) - \frac{1}{{N + 1}}$

If you cannot what is going on there, then you need more help than you can get here,