series limit

**gilyos** · February 1st 2010, 11:07 AM

$a_{n} = \frac{1}{n+1} + \frac{1}{n+2} + ... \frac{1}{2n}$

what is the limit ?

**Henryt999** · February 1st 2010, 11:58 AM

Im trying to learn the same things so make sure to check my answer with someone else here but what I noticed is that the difference between each term is 1. Doesn´t that imply that this is an aritmetic series?
There is a formula for that and that is $\sum{a_1}+{a_2}....+a_n = \frac{n(a_1+a_n)}{2}$

**girdav** · February 1st 2010, 11:59 AM

We have $a_{n+1}-a_n = \frac 1{2n+1}+\frac 1{2n+2}-\frac 1{n+1} <br /> = \frac 1 {\left(2n+1\right)\left(2n+2\right)}$ so you can write
$a_n$ as a sum.
Then you will have to study the convergence of a serie.

**Krizalid** · February 1st 2010, 01:26 PM

http://www.mathhelpforum.com/math-he...tml#post430479

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