# Sum of convergent series

• Jul 14th 2013, 06:59 AM
minneola24
Sum of convergent series
How do I find the sum of this?

I got the sum of 2/3 ^n because its a Geometric sequence.

But the other part is tricky. I separated into two partial fractions but am stuck.Attachment 28824
• Jul 14th 2013, 07:34 AM
Plato
Re: Sum of convergent series
Quote:

Originally Posted by minneola24
How do I find the sum of this?
I got the sum of 2/3 ^n because its a Geometric sequence.
But the other part is tricky. I separated into two partial fractions but am stuck.Attachment 28824

That is a collapsing sum: $\displaystyle \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{(n + 1)(n + 2)}}} \right)} = \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{(n + 1)}} - \frac{1}{{(n + 2)}}} \right)}$

Any partial sum looks like $\displaystyle {S_K} = \sum\limits_{n = 1}^N {\left( {\frac{1}{{(n + 1)}} - \frac{1}{{(n + 2)}}} \right)} = \frac{1}{2} - \frac{1}{{N + 2}}$ thus $\displaystyle \left(S_N\right)\to~?$.
• Jul 14th 2013, 07:44 AM
minneola24
Re: Sum of convergent series
How did you get to .5-1/(n+1)?

How do i complete this partial sum?
Thanks
• Jul 14th 2013, 07:53 AM
Plato
Re: Sum of convergent series
Quote:

Originally Posted by minneola24
How did you get to .5-1/(n+1)?
How do i complete this partial sum?

Do you know what it actually means for a infinite series to converge?
What do partial sums have to do with?

Do you understand how $\displaystyle {S_N} = \sum\limits_{n = 1}^N {\left( {\frac{1}{{(n + 1)}} - \frac{1}{{(n + 2)}}} \right)} = \frac{1}{2} - \frac{1}{{N + 2}}$ actually works?

If not, there is no point in your trying the question.