Infinite series for partial sums of square roots.

Question

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]

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 01 '14 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 01 '14 at 14:43

score 7 · Accepted Answer · answered Mar 01 '14 at 14:45

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

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

score 6 · Answer 2 · answered Mar 01 '14 at 15:00

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

Once more, a beautiful and elegant answer ! – Claude Leibovici Mar 01 '14 at 16:03 — Claude Leibovici, Mar 01 '14 at 16:03

score 3 · Answer 3 · answered Mar 01 '14 at 14:59

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.

score 1 · Answer 4 · answered Mar 01 '14 at 19:50

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.

Infinite series for partial sums of square roots.

4 Answers4