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

what is the limit ?

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- Feb 1st 2010, 11:07 AMgilyosseries limit
$\displaystyle a_{n} = \frac{1}{n+1} + \frac{1}{n+2} + ... \frac{1}{2n} $

what is the limit ? - Feb 1st 2010, 11:58 AMHenryt999Hii
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 $\displaystyle \sum{a_1}+{a_2}....+a_n = \frac{n(a_1+a_n)}{2}$ - Feb 1st 2010, 11:59 AMgirdav
We have$\displaystyle a_{n+1}-a_n = \frac 1{2n+1}+\frac 1{2n+2}-\frac 1{n+1}

= \frac 1 {\left(2n+1\right)\left(2n+2\right)}$ so you can write

$\displaystyle a_n$ as a sum.

Then you will have to study the convergence of a serie. - Feb 1st 2010, 01:26 PMKrizalid