evaluate $\sum_1^\infty \frac{1}{(n \sqrt{n+1} + (n+1)\sqrt{n})}$

Question

How can I evaluate the infinite sum $\sum_1^\infty \frac{1}{(n \sqrt{n+1} + (n+1)\sqrt{n})}$? I try to change that into the form $\sum_1^\infty \frac{1}{(\sqrt{n} + \frac{n}{n+1}\sqrt{n+1})}-\frac{1}{(\sqrt{n+1} + \frac{n+1}{n}\sqrt{n})}$ and try to reduce the terms. But it seems not valid. I'm thinking of using some fourier transform techniques, any suggestions?

Answer 1

Since $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}}{a-b}$,$$\frac{1}{\sqrt{(n+1)^2n}+\sqrt{n^2(n+1)}}=\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}=u_n-u_{n+1}$$with $u_n:=\frac{1}{\sqrt{n}}$, so the series telescopes to $u_1=1$.

Answer 2

J.G. was too fast for me...

My approach: I assumed there is no easy solution, so I tried to get an approximate solution. The values summed are roughly n^(-3/2). Even if you can't find an exact telescopic series, if you find one that is close to n^(-3/2) that gives you something you can easily sum up, plus a remainder that is a lot smaller, and then you can repeat the same approach.

Obviously with f(n) = -n^(-1/2), we have f(n+1) - f(n) ≈ n^(-3/2). So a guess is that with f(n) = -(n+c)^(1/2) for the right constant c, f(n+1) - f(n) should be a good approximation. How to find c? I used a spreadsheet to calculate some values. To my surprise, c = 0 gave an exact solution!

Now this isn't very elegant, but it helps with slightly different sums. For example if the divisor was n sqrt(n+2) + (n+1) sqrt(n), f(n) = -(n + 1/6)^(-1/2) gives a very nice approximation to a telescopic series; f(n+1) - f(n) has an error around 8 / n^3.5 when n is large. (I do prefer solutions that are not useful for one single problem only, but can be used for similar problems as well).