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Can you prove these infinite series for partial sums of square roots?

$$\sqrt{1}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$

$$\sqrt{1}+\sqrt{2}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)$$

$$\sqrt{1}+\sqrt{2}+\sqrt{3}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}-\frac{3}{\sqrt{n+3}}\right)$$

$$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}=\sum _{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}-\frac{4}{\sqrt{n+4}}\right)$$

$$\cdots$$

And is there some easy cancellation that I have missed on the right hand side?

Mathematica:

Clear[s, i, n, j]
s = 1/2; 
i = 1; 
j = 0; 
Sum[1/(n + 0)^s - 1/(n + 1)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s - 2/(n + 2)^s + 1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s - 3/(n + 3)^s + 1/(n + 0)^s, 
   {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s - 4/(n + 4)^s + 
     1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
N[Accumulate[Sqrt[Range[4]]], 20]
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3  
Try writing out the terms. –  John Habert Mar 1 at 14:37
    
The problem is that the sequence of numbers on the right hand side within the sum, is so long that I can't spot the possible cancellations. –  Mats Granvik Mar 1 at 14:42
    
For the first one at least, it should be very obvious after about the first 3 terms or so. The others will have similar style cancellation though a bit different. –  John Habert Mar 1 at 14:43

4 Answers 4

up vote 6 down vote accepted

Let's see how we find the first sum which's known as telescoping sum and the other sums are almost the same: the idea is to change the index and then cancel most of the terms

$$\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \frac{1}{\sqrt{n}}-\sum\limits_{n=2}^{N+1}\frac{1}{\sqrt{n}}\\=\lim_{N\to\infty}\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{N+1}}=\frac{1}{\sqrt{1}}$$

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Excellent, Sami! –  amWhy Mar 2 at 14:33

$$ \begin{align} \sum_{n=1}^\infty\left(\left(\sum_{k=0}^{m-1}\frac1{\sqrt{n+k}}\right)-\frac{m}{\sqrt{n+m}}\right) &=\lim_{N\to\infty}\sum_{n=1}^N\sum_{k=0}^{m-1}\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\ &=\lim_{N\to\infty}\sum_{k=0}^{m-1}\sum_{n=1}^N\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\ &=\lim_{N\to\infty}\sum_{k=0}^{m-1}\left(\sum_{n=k+1}^{k+N}\frac1{\sqrt{n}}-\sum_{n=m+1}^{m+N}\frac1{\sqrt{n}}\right)\\ &=\lim_{N\to\infty}\sum_{k=0}^{m-1}\left(\sum_{n=k+1}^m\frac1{\sqrt{n}}-\sum_{n=k+N+1}^{m+N}\frac1{\sqrt{n}}\right)\\ &=\sum_{k=0}^{m-1}\sum_{n=k+1}^m\frac1{\sqrt{n}}\\ &=\sum_{n=1}^m\sum_{k=0}^{n-1}\frac1{\sqrt{n}}\\ &=\sum_{n=1}^m\sqrt{n} \end{align} $$

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Once more, a beautiful and elegant answer ! –  Claude Leibovici Mar 1 at 16:03

You have a very elegant demonstration given by Sami Ben Romdhane.

If, instead of summing to $\infty$, you sum to $m$, you will get the following formulas for the different sums $$\sqrt{1}-\frac{1}{\sqrt{m+1}}$$ $$\sqrt{1}+\sqrt{2}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}-2 H_{m+2}^{\left(\frac{1}{2}\right)}$$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}+H_{m+2}^{\left(\frac{1}{2}\right)}-3 H_{m+3}^{\left(\frac{1}{2}\right)}$$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}+H_{m+2}^{\left(\frac{1}{2}\right)}+H_{m+3}^{\left(\frac{1}{2}\right)}-4 H_{m+4}^{\left(\frac{1}{2}\right)}$$ Now, push $m$ to $\infty$ and use the properties of the harmonic numbers.

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For the first one, $$\begin{align*} \sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) &= \sum_{n=1}^\infty \frac{1}{\sqrt{n}}-\sum_{n=1}^\infty\frac{1}{\sqrt{n+1}} \\ &= \sum_{n=1}^\infty \frac{1}{\sqrt{n}}-\sum_{n=2}^\infty\frac{1}{\sqrt{n}} \\ &= \frac{1}{\sqrt{1}}\\ &= \sqrt{1}. \end{align*}$$

For the second, $$\begin{align*} \sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right) &=\sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) +\sum_{n=1}^\infty\left(\frac{2}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)\\ &= \sqrt{1} + 2\sum_{n=2}^\infty\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\\ &= \sqrt{1} + \frac{2}{\sqrt{2}} \\ &= \sqrt{1} + \sqrt{2}. \end{align*}$$

And so on. It would probably be faster to set up an induction using the ideas from the second case than to prove the third and fourth cases individually.

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