1/[ n*sqrt(n+1) + (n+1)*sqrt(n) ] finding sum of this serie

Hello,

I am trying to solve this series:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n\sqrt(n+1)+(n+1)\sqrt (n)}$

I have tried partial decomposition (telescoping series) process but it results very complicated and incorrect and then I get stack.

Have anyone an idea how the partial decompostion should be done so I get a result that the terms start to canceling them self. I don't want to bother anymore, just that part is enough and I will continue with the rest of the solution.

Thank you everyone!

Re: 1/[ n*sqrt(n+1) + (n+1)*sqrt(n) ] finding sum of this serie

You probably need to start by rationalising the denominator.

$\displaystyle \displaystyle \begin{align*} \frac{1}{n\sqrt{n + 1} + (n + 1)\sqrt{n}} &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{[n\sqrt{n + 1} + (n + 1)\sqrt{n}][n\sqrt{n + 1} - (n + 1)\sqrt{n}]} \\ &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{n^2(n + 1) - n(n + 1)^2}\\ &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{n(n + 1)[n - (n + 1)]} \\ &= \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{n(n + 1)} \\ &= \frac{(n + 1)\sqrt{n } - n\sqrt{n + 1}}{(\sqrt{n})^2(\sqrt{n + 1})^2}\end{align*}$

Now try the Partial Fractions decomposition.

Re: 1/[ n*sqrt(n+1) + (n+1)*sqrt(n) ] finding sum of this serie

$\displaystyle \begin{aligned}& \sum_{1 \le n \le k}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} = \sum_{1 \le n \le k}\bigg(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\bigg) = \sum_{1 \le n \le k}\frac{1}{\sqrt{n}}-\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}} \\& = \sum_{1 \le n+1 \le k}\frac{1}{\sqrt{n+1}}-\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}} = \sum_{0 \le n \le k-1}\frac{1}{\sqrt{n+1}}-\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}} \\& = 1-\frac{1}{\sqrt{k+1}}+\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}}-\sum_{1 \le n \le k}\frac{1}{\sqrt{n+1}} = 1-\frac{1}{\sqrt{k+1}}.\end{aligned}$

$\displaystyle \begin{aligned}\therefore \sum_{n \ge 1}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} = \lim_{k \to +\infty}\sum_{1 \le n \le k}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} = \lim_{k \to +\infty}\bigg(1-\frac{1}{\sqrt{k+1}}\bigg) = 1.\end{aligned}$

Re: 1/[ n*sqrt(n+1) + (n+1)*sqrt(n) ] finding sum of this serie

Quote:

Originally Posted by

**Prove It** You probably need to start by rationalising the denominator.

$\displaystyle \displaystyle \begin{align*} \frac{1}{n\sqrt{n + 1} + (n + 1)\sqrt{n}} &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{[n\sqrt{n + 1} + (n + 1)\sqrt{n}][n\sqrt{n + 1} - (n + 1)\sqrt{n}]} \\ &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{n^2(n + 1) - n(n + 1)^2}\\ &= \frac{n\sqrt{n + 1} - (n + 1)\sqrt{n}}{n(n + 1)[n - (n + 1)]} \\ &= \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{n(n + 1)} \\ &= \frac{(n + 1)\sqrt{n } - n\sqrt{n + 1}}{(\sqrt{n})^2(\sqrt{n + 1})^2}\end{align*}$

Now try the Partial Fractions decomposition.

What's with the last step? From the second last step, I think you can separate it into two fractions:

$\displaystyle \begin{aligned}\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)} = \frac{(n+1)\sqrt{n}}{n(n+1)}-\frac{n\sqrt{n+1}}{n(n+1)} = \frac{\sqrt{n}}{{n}}-\frac{\sqrt{n+1}}{n+1} = \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\end{aligned}.$

EDIT: now I realised that the last step might have been a hint towards breaking it into two fractions, as in above!