Infinite series for partial sums of square roots. 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]

 A: 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
$$\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \frac{1}{\sqrt{n}}-\sum\limits_{n=2}^{N+1}\frac{1}{\sqrt{n}}\\=\lim_{N\to\infty}\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{N+1}}=\frac{1}{\sqrt{1}}$$
A: $$
\begin{align}
\sum_{n=1}^\infty\left(\left(\sum_{k=0}^{m-1}\frac1{\sqrt{n+k}}\right)-\frac{m}{\sqrt{n+m}}\right)
&=\lim_{N\to\infty}\sum_{n=1}^N\sum_{k=0}^{m-1}\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\
&=\lim_{N\to\infty}\sum_{k=0}^{m-1}\sum_{n=1}^N\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\
&=\lim_{N\to\infty}\sum_{k=0}^{m-1}\left(\sum_{n=k+1}^{k+N}\frac1{\sqrt{n}}-\sum_{n=m+1}^{m+N}\frac1{\sqrt{n}}\right)\\
&=\lim_{N\to\infty}\sum_{k=0}^{m-1}\left(\sum_{n=k+1}^m\frac1{\sqrt{n}}-\sum_{n=k+N+1}^{m+N}\frac1{\sqrt{n}}\right)\\
&=\sum_{k=0}^{m-1}\sum_{n=k+1}^m\frac1{\sqrt{n}}\\
&=\sum_{n=1}^m\sum_{k=0}^{n-1}\frac1{\sqrt{n}}\\
&=\sum_{n=1}^m\sqrt{n}
\end{align}
$$
A: 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.
A: 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.
