On the sum of the square roots of the first n natural numbers
Journal of the Indian Mathematical Society, VII, 1915, 173 – 175

1. Let \begin{align*} \phi_1(n) = &\sqrt{1} + \sqrt{2} + \sqrt{3} + \cdots + \sqrt{n} - \left(C_1 + \mbox{$\frac{2}{3}$} n \sqrt{n} + \half \sqrt{n}\right)\\ &- \mbox{$\frac{1}{6}$} \sum^{\nu=\infty}_{\nu=0} \{\sqrt{(n + \nu)} + \sqrt{(n+ \nu + 1)}\}^{-3}, \end{align*} where $C_1$ is a constant such that $\phi_1 (1)= 0.$ Then we see that \begin{align*} \phi_1 (n)-\phi_1 (n+1) = &- \sqrt{(n+1)} + \left [\mbox{$\frac{2}{3}$} (n+1) \sqrt{(n+1)} + \half \sqrt{(n+1)}\right]\\ &- (\mbox{$\frac{2}{3}$} n \sqrt{n} + \half \sqrt{n}) + \mbox{$\frac{1}{6}$} \{\sqrt{n}-\sqrt{(n+1)}\}^3 = 0. \end{align*}

But $\phi_1(1) = 0.$ Hence $\phi_1(n) = 0$ for all values of $n$. That is to say

\begin{align} & \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + \cdots + \sqrt{n} = C_1 + \mbox{$\frac{2}{3}$} n \sqrt{n} + \half \sqrt{n} + \mbox{$\frac{1}{6}$} \left[ \left\{\sqrt{n} + \sqrt{(n+1)}\right\}^{-3}\right.\notag\\ &\hspace{3.6cm} \left .+ \left\{\sqrt{(n+1)} + \sqrt{(n+2)}\right\}^{-3} +\left\{\sqrt{(n+2)} + \sqrt{(n+3)}\right\}^{-3} + \cdots \right]. \end{align}
But it is known that
$$C_1 = \frac{1}{4 \pi} \left(\frac{1}{1\sqrt{1}} + \frac{1}{2\sqrt{2}} + \frac{1}{3\sqrt{3}} + \cdots \right).$$
Putting $n=1$ in (1) and using (2), we obtain
\begin{align} & 2 \pi \left\{ \frac{1}{(\sqrt{1})^3} + \frac{1}{(\sqrt{1} + \sqrt{2})^3} + \frac{1}{(\sqrt{2} + \sqrt{3})^3} + \frac{1}{(\sqrt{3} + \sqrt{4})^3} + \cdots \right\}\notag\\ &\hspace{3.9cm} = 3 \left\{\frac{1}{(\sqrt{1})^3} + \frac{1}{(\sqrt{2})^3} + \frac{1}{(\sqrt{3})^3} + \frac{1}{(\sqrt{4})^3} + \cdots \right\}. \end{align}

2. Again let \begin{align*} \phi_2 (n) = 1 \sqrt{1} + 2\sqrt{2} \ldots + & n \sqrt{n} - \left(C_2 + \mbox{$\frac{2}{5}$} n^2 \sqrt{n} + \half n \sqrt{n} + \mbox{$\frac{1}{8}$} \sqrt{n}\right)\\ & - \mbox{$\frac{1}{40}$} \sum^{\nu=\infty}_{\nu=0} [\sqrt{(n+\nu)} + \sqrt{(n+\nu+1)}]^{-5}, \end{align*} where $C_2$ is a constant such that $\phi_2 (1) = 0.$ Then we have \begin{align*} \phi_2 (n) - \phi_2 (n+1) & = - (n+1) \sqrt{(n+1)}\\ &\quad + \{\mbox{$\frac{2}{5}$} (n+1)^2 \sqrt{(n+1)} + \half (n+1) \sqrt{(n+1)} + \mbox{$\frac{1}{8}$} \sqrt{(n+1)}\} \\ &\quad - \{\mbox{$\frac{2}{5}$} n^2 \sqrt{n} + \half n \sqrt{n} + \mbox{$\frac{1}{8}$} \sqrt{n} \} +\mbox{$\frac{1}{40}$} \{\sqrt{n} - \sqrt{(n+1)}\}^5 = 0. \end{align*} But $\phi_2 (1) = 0.$ Hence $\phi_2 (n) = 0.$ In other words

\begin{align} 1 \sqrt{1} + 2 \sqrt{2} + 3 \sqrt{3} & + \cdots n \sqrt{n} = C_2 + \mbox{$\frac{2}{5}$} n^2 \sqrt{n} + \half n \sqrt{n} + \mbox{$\frac{1}{8}$} \sqrt{n} + \mbox{$\frac{1}{40}$}\left[\{\sqrt{n} + \sqrt{(n+1)}\}^{-5}\right.\notag\\ & \left. +\{\sqrt{(n+1)} + \sqrt{(n+2)}\}^{-5} + \{\sqrt{(n+2)} + \sqrt{(n+3)}\}^{-5} + \cdots \right]. \end{align}
But it is known that
$$C_2 = -\frac{3}{16 \pi^2} \left(\frac{1}{1^2 \sqrt{1}} + \frac{1}{2^2 \sqrt{2}} + \frac{1}{3^2 \sqrt{3}} + \cdots \right).$$
It is easy to see from (4) and (5) that
\begin{align} & 2 \pi^2 \left\{\frac{1}{(\sqrt{1})^5} + \frac{1}{(\sqrt{1} + \sqrt{2})^5} + \frac{1}{(\sqrt{2}+\sqrt{3})^5} + \frac{1}{(\sqrt{3} + \sqrt{4})^5} + \cdots \right\}\notag \\ & \hspace{3cm} = 15 \left\{\frac{1}{(\sqrt{1})^5} + \frac{1}{(\sqrt{2})^5} + \frac{1}{(\sqrt{3})^5} + \frac{1}{(\sqrt{4})^5} + \cdots \right\}. \end{align}

3. The corresponding results for higher powers are not so neat as the previous ones. Thus for example

\begin{align} & 1^2 \sqrt{1} + 2^2 \sqrt{2} + 3^2 \sqrt{3} + \cdots + n^2 \sqrt{n} = C_3 +\sqrt{n}(\mbox{$\frac{2}{7}$} n^3 + \half n^2 +\mbox{$\frac{5}{24}$}n)\notag\\ &\qquad -\mbox{$\frac{1}{96}$} [\{\sqrt{n} + \sqrt{(n+1)}\}^{-3} + \{\sqrt{(n+1)} + \sqrt{(n+2)}\}^{-3} + \cdots ] \notag\\ &\qquad +\mbox{$\frac{1}{224}$}\left [\{\sqrt{n} + \sqrt{(n+1)}\}^{-7} + \{\sqrt{(n+1)} + \sqrt{(n+2)}\}^{-7}\right.\notag\\ &\qquad\quad \left.+ \{\sqrt{(n+2)} + \sqrt{(n+3)}\}^{-7} + \cdots\right]; \end{align}
\begin{align} & 1^3 \sqrt{1} + 2^3 \sqrt{2} + \cdots + n^3 \sqrt{n} = C_4 + \sqrt{n} (\mbox{$\frac{2}{9}$} n^4 + \half n^3 + \mbox{$\frac{7}{24}$} n^2 - \mbox{$\frac{7}{384}$})\notag\\ &\quad -\mbox{$\frac{1}{192}$} \left[\{\sqrt{n} + \sqrt{(n+1)}\}^{-5} + \{\sqrt{(n+1)} +\sqrt{(n+2)}\}^{-5} + \cdots\right]\notag\\ &\quad + \mbox{$\frac{1}{1152}$} [\{\sqrt{n} +\sqrt{(n+1)}\}^{-9} + \{\sqrt{(n+1)} +\sqrt{(n+2)}\}^{-9} + \cdots]; \end{align}
and so on.

The constants $C_3, C_4, \ldots$ can be ascertained from the well-known result that the constant in the approximate summation of the series $1^{r-1} + 2^{r-1} + 3^{r-1} + \cdots + n^{r-1}$ is

$$\frac{2\Gamma (r)}{(2\pi)^r} \left(\frac{1}{1^r} + \frac{1}{2^r} + \frac{1}{3^r} + \frac{1}{4^r} + \cdots \right) \cos \half \pi r,$$
provided that the real part of $r$ is greater than 1.

4. Similarly we can shew, by induction, that

\begin{align} \frac{1}{\sqrt{1}} & + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} = C_0 + 2\sqrt{n} + \frac{1}{2\sqrt{n}}\notag\\ &- \half \left\{\frac{\{\sqrt{n} + \sqrt{(n+1)}\}^{-3}}{\sqrt{\{n(n+1)\}}} + \frac{\{\sqrt{(n+1)} + \sqrt{(n+2)^{-3}}\}}{\sqrt{\{(n+1)(n+2)}\}} + \cdots \right\}, \end{align}
The value of $C_0$ can be determined as follows: from (10) we have
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{(2n)}} - 2 \sqrt{(2n)} \to C_0,$$
as $n \to \infty$. Also
$$2 \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{6}} + \cdots + \frac{1}{\sqrt{(2n)}}\right) - 2 \sqrt{(2n)} \to C_0 \sqrt{2},$$
as $n \to \infty$.

Now subtracting (12) from (11) we see that $$\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \cdots - \frac{1}{\sqrt{(2n)}} \to C_0 (1- \sqrt{2}), \:\:{\rm as}\:\: n \to \infty.$$ That is to say

$$C_0 = - (1+ \sqrt{2}) \left(\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \cdots \right).$$
We can also shew, by induction, that
\begin{align} \sqrt{1}+\sqrt{2}+\sqrt{3} & +\cdots +\sqrt{n}=C_1 + \mbox{$\frac{2}{3}$} n \sqrt{n} + \half \sqrt{n}+\frac{1}{24\sqrt{n}}\notag\\ & -\mbox{$\frac{1}{24}$} \left[\frac{\{\sqrt{n} + \sqrt{(n+1)}\}^{-5}}{\sqrt{\{n(n+1)\}}} + \frac{\{\sqrt{(n+1)} + \sqrt{(n+2)}\}^{-5}}{\sqrt{\{(n+1) (n+2)\}}} + \cdots \right]. \end{align}

The asymptotic expansion of $\sqrt{1} + \sqrt{2} + \sqrt{3} + \cdots + \sqrt{n}$ for large values of $n$ can be shewn to be

$$C_1 + \mbox{\frac{2}{3}} n \sqrt{n} + \half \sqrt{n} + \frac{1}{\sqrt{n}} \left(\mbox{\frac{1}{24}} -\frac{1}{1920n^2} + \frac{1}{9216 n^4} - \cdots \right),$$
by using the Euler-Maclaurin sum formula.