# Limit of sequence

1. Dec 14, 2009

### llursweetiell

1. The problem statement, all variables and given/known data
consider the sequence {an} where an= 1/(n+1) + ...+ 1/2n. find its limit.

2. Relevant equations
the hint given is using riemann sum.

3. The attempt at a solution
we know that since it is increasing and bounded above by one, the sequence converges. I'm not sure where to go from there, especially to use the riemann sum.

2. Dec 14, 2009

### snipez90

Yes this question basically asks you if you know how Riemann or Darboux sums work. Hint: consider the function f defined by f(x) = 1/(1+x) on [0,1]. Choose the most standard partition of an interval you can think of and find out how this relates to your sum.

3. Dec 14, 2009

### llursweetiell

I'm thinking that the partition would be P= 0=x0<x1<x2<...,xn=1 with 1/n partitions. Not sure about how this would relate to the sum that I have in the problem?

4. Dec 14, 2009

### snipez90

Yes, that is the right partition if you meant that there are n subintervals each of length 1/n. Now rewrite the original sum as

$$\sum_{k = 1}^{n}\frac{1}{n+k} = \sum_{k = 1}^{n}\left(\frac{1}{1+\frac{k}{n}}\right)\cdot\frac{1}{n}.$$

Finally, consider the upper and lower sums.

5. Dec 14, 2009

### llursweetiell

Here is what i've come up with:
let f(x)=x/(x+1)
let P be the partition {x0, x1, ...xn} where xi=i/n and ti=xi and the mesh is 1/n=xi-xi-1 (i and i-1 are subscripts)

then S(P,f)=SUM: f(ti)(xi-xi-1) = SUM: 1/(n+1) * 1/n, from i=1 to n which is the sequence an.
since S(p,f) converges to the integral of 1/(x+1)dx, from 0 to 1, an converges to this as well. so the limit of an is the value of the integral, which is ln2.

Is that where you were going with it?

Last edited: Dec 14, 2009