# Math Help - need help with series limit

1. ## need help with series limit

i know it's not very complicated, i just need help understanding this one

$lim(1/1*2+1/2*3+...+1/(n*(n+1)))= ?$
n->inf

i already know the answer is 1, and i can see how the fractions are adding up to something that is <1 but i don't know how to get to that conclusion

sorry i still don't know how to use latex, it will be different on my next posts

thanks!

2. ## Re: need help with series limit

I'm thinking I'd resort to a partial fraction decomposition of the nth term.

3. ## Re: need help with series limit

can you elaborate on that?..

4. ## Re: need help with series limit

Did you do it?

$\frac{1}{n\cdot(n+1)}\;=\;\frac{?}{n}+\frac{??}{n+ 1}$

5. ## Re: need help with series limit

no because i don't understand what your'e getting at

i know that the solution for what your'e asking is

$1/n - 1/(n+1)$

but what next?

6. ## Re: need help with series limit

Originally Posted by idom87
no because i don't understand what your'e getting at
Sometimes this is just necessary:
$\sum\limits_{n - 1}^N {\frac{1}{{n(n + 1)}}} = \sum\limits_{n - 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right] = 1 - \frac{1}{{N + 1}}}$

7. ## Re: need help with series limit

Originally Posted by Plato
Sometimes this is just necessary:
$\sum\limits_{n - 1}^N {\frac{1}{{n(n + 1)}}} = \sum\limits_{n - 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right] = 1 - \frac{1}{{N + 1}}}$
so the limit of $1 - \frac{1}{{N + 1}}$
is 1, ok

but i still can't wrap my head around this part
$\sum\limits_{n - 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right] = 1 - \frac{1}{{N + 1}}$
can you explain why?

8. ## Re: need help with series limit

Originally Posted by idom87
so the limit of $1 - \frac{1}{{N + 1}}$
is 1, ok

but i still can't wrap my head around this part
$\sum\limits_{n - 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right] = 1 - \frac{1}{{N + 1}}$
can you explain why?
Let $N=4$ actually write the four terms.
See how all but the first and the last 'collapse' .
These are called collapsing sums .

9. ## Re: need help with series limit

thanks i got that after i wrote down your example

what about if i have a more complex nth term?..
like this one

$lim (1/(1*4) + 1/(4*7) + ... + 1/((3n-2)(3n+1)))$
n->inf

?

10. ## Re: need help with series limit

Same, split in partial fractions:
$\frac{1}{(3n-2)\cdot (3n+1)}=\frac{A}{3n-2}+\frac{B}{3n+1}=...$

11. ## Re: need help with series limit

how do i find A and B?

all i know is that

A(3n+1)+B(3n-2)=1

12. ## Re: need help with series limit

Originally Posted by idom87
how do i find A and B?
A(3n+1)+B(3n-2)=1
Have a look at this webpage.
Be sure to click the show steps button.

13. ## Re: need help with series limit

damn i feel stupid now
and rusty
awsome page

14. ## Re: need help with series limit

Originally Posted by idom87
damn i feel stupid now
and rusty
awsome page
You should really explore that website.