# Limit of sequence

#### 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.

#### 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.

#### 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?

#### 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.

#### 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?

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