Summation of a certain series
Messenger of Mathematics, XLIV, 1915, 157 – 160

1. Let \begin{eqnarray*} \Phi (s, x) &=& \sum^{n=\infty}_{n=0} \{\sqrt{(x+n)} + \sqrt{(x+n+1)}\}^{-s}\\ &=& \sum^{n=\infty}_{n=0} \{\sqrt{(x+n+1)} - \sqrt{(x+n)}\}^s. \end{eqnarray*} The object of this paper is to give a finite expression of $\Phi (s,0)$ in terms of Riemann $\zeta$-functions, when $s$ is an odd integer greater than 1. Let $\zeta(s,x)$, where $x \gt 0$, denote the function expressed by the series $$x^{-s} + (x+1)^{-s} + (x+2)^{-s} + \cdots,$$ and its analytical continuations. Then

$$\zeta (s, 1) = \zeta (s), \zeta (s, \half ) = (2^s - 1) \zeta (s),$$
where $\zeta(s)$ is the Riemann $\zeta$-function;
$$\zeta (s, x) - \zeta(s, x + 1) = x^{-s};$$
\begin{eqnarray} \left.\begin{array}{ll} 1^s + 2^s + 3^s + \cdots + n^s = \zeta (-s) - \zeta (-s, n+1), \\ 1^s + 3^s+ 5^s + \cdots + (2n-1)^s = (1-2^s) \zeta (-s) - 2^s \zeta (-s, n + \half )\end{array}\right\} , \end{eqnarray}
if $n$ is a positive integer; and
\begin{eqnarray} \lim_{x \to \infty} \left\{ \zeta (s, x) - \half x^{-s} + \left(\frac{x^{1-s}}{1-s} - B_2 \frac{s}{2!} x^{-s-1} + B_4 \frac{s(s+1)(s+2)} {4!} x^{-s-3}\right.\right.\nonumber\\ \left.\left.- B_6 \frac{s(s+1)(s+2)(s+3)(s+4)}{6!} x^{-s-5} + \cdots \mbox{ to } n \mbox{ terms }\right)\right\} = 0, \end{eqnarray}
if $n$ is a positive integer, $-(2n-1) \lt s \lt 1,$ and $B_2 = \frac{1}{6}, B_4 = \frac{1}{30}$, $B_6 = \frac{1}{42}$, $B_8= \frac{1}{30}, \ldots$, are Bernoulli's numbers. Suppose now that $$\Psi (x) = 6 \zeta (- \half , x) + (4x - 3) \sqrt{x} + \Phi (3, x).$$ Then from (2) we see that \begin{equation*} \Psi (x) - \Psi(x+1) = 6 \sqrt x + (4x-3)\sqrt{x} -(4x+1)\sqrt{(x+1)} + \{\sqrt{(x+1)} - \sqrt{x}\}^3 = 0; \end{equation*} and from (4) that $\Psi (x) \to 0$ as $x \to \infty$. It follows that $\Psi (x)=0$. That is to say,
$$6 \zeta (-\half , x) + (4x - 3) \sqrt{x} + \Phi (3,x) = 0.$$
Similarly, we can shew that
$$40 \zeta (-\frac{3}{2}, x) + (16 x^2 - 20x + 5) \sqrt{x} + \Phi (5, x) = 0.$$

2. Remembering the functional equation satisfied by $\zeta(s)$, viz.,

$$\zeta (1-s) = 2 (2 \pi)^{-s} \Gamma (s) \zeta(s) \cos \half \pi s,$$
we see from (3) and (5) that
$$\sqrt{1} + \sqrt{2} + \sqrt{3} + \cdots + \sqrt{n} = \frac{2}{3} n^{\frac{3}{2}} + \half \sqrt{n} - \frac{1}{4 \pi} \zeta (\frac{3}{2}) + \frac{1}{6} \Phi (3, n);$$
and
\begin{eqnarray} \sqrt{1} &+& \sqrt{3} + \sqrt{5} + \cdots + \sqrt{(2n-1)} \nonumber \\ &=& \frac{1}{3} (2n - 1)^{\frac{3}{2}} + \half \sqrt{(2n-1)} + \frac{\sqrt{2-1}}{4 \pi} \zeta (\frac{3}{2}) + \frac{1}{3\sqrt{2}} \Phi (3, n - \half ). \end{eqnarray}
Similarly from (6), we have
\begin{eqnarray} 1 \sqrt{1} &+& 2 \sqrt{2} + 3 \sqrt{3} + \cdots + n\sqrt{n} \nonumber\\ &=& \frac{2}{5} n^{\frac{5}{2}} + \half n^{\frac{3}{2}} + \frac{1}{8} \sqrt{n} - \frac{3}{16 \pi^2} \zeta(\frac{5}{2}) + \frac{1}{40} \Phi (5, n); \end{eqnarray}
and
\begin{eqnarray} 1 \sqrt{1} &+& 3 \sqrt{3} + 5 \sqrt{5} + \cdots + (2n-1) \sqrt{(2n-1)} \nonumber \\ &=& \frac{1}{5} (2n-1)^{\frac{5}{2}} + \half (2n-1)^{\frac{3}{2}} + \frac{1}{4} \sqrt{(2n-1)} \nonumber \\ &+& \frac{3(2\sqrt{2-1}}{16 \pi^2} \zeta (\mbox{$\frac{5}{2}$}) + \frac{1}{10 \sqrt{2}}\Phi (5, n - \half ). \end{eqnarray}
It also follows from (5) and (6) that
\begin{eqnarray} && \sqrt{(a+d)} + \sqrt{(a+2d)} + \sqrt{(a+3d)} + \cdots + \sqrt{(a+nd)}\nonumber \\ &=& C + \frac{2}{3d} (a+nd)^{\frac{3}{2}} + \half \sqrt{(a+nd)} + \frac{1}{6} \sqrt{d} \Phi (3, n + a/d); \end{eqnarray}
and
\begin{eqnarray} && (a+d)^{\frac{3}{2}} + (a+2d)^{\frac{3}{2}} + (a+3d)^{\frac{3}{2}} + \cdots + (a+nd)^{\frac{3}{2}} \nonumber \\ &=& C' + \frac{2}{5d} (a+nd)^{\frac{5}{2}} + \half (a+nd)^{\frac{3}{2}} + \frac{1}{8} d \sqrt{(a+nd)} + \frac{1}{40} d \sqrt{d} \Phi (5, n + a/d), \end{eqnarray}
where $C$ and $C'$ are independent of $n$. Putting $n=1$ in (8) and (10), we obtain
$$\Phi (3,0) = \frac{3}{2 \pi} \zeta~(\frac{3}{2}), \Phi(5,0) = \frac{15}{2 \pi^2} \zeta (\frac{5}{2}).$$

3. The preceding results may be generalised as follows. If $s$ be an odd integer greater than 1, then

\begin{eqnarray} \Phi (s, x) &+& \half \{\sqrt{x} + \sqrt{(x-1)}\}^s + \half \{\sqrt{x} - \sqrt{(x-1)}\}^s \nonumber\\ \quad\quad &+& \frac{s}{1!} 2^{s-2} \zeta (1-\half s, x) + \frac{s (s-4)(s-5)}{3!} 2^{s-6} \zeta ( 3 - \half s, x) \nonumber \\ \quad\quad &+& \frac{s(s-6)(s-7) (s-8)(s-9)}{5!} 2^{s-10} \zeta ( 5 -\half s,x)\nonumber \\ \quad\quad &+& \frac{s(s-8)(s-9)(s-10)(s-11)(s-12)(s-13)}{7!} 2^{s-14} \nonumber \\ && \hspace{4cm}\times \zeta (7 - \half s, x) + \cdots \mbox{ to } [\mbox{$\frac{1}{4}$} (s+1)] \mbox{ terms }=0, \end{eqnarray}
where $[x]$ denotes, as usual, the integral part of $x$. This can be proved by induction, using the formula
\begin{eqnarray} \{\sqrt{x} &+& \sqrt{(x \pm 1)}\}^s + \{\sqrt{x} - \sqrt{(x \pm 1)}\}^s \nonumber \\ &=& (2 \sqrt{x})^s \pm \frac{s}{1!} (2 \sqrt{x})^{s-2} + \frac{s(s-3)}{2!} (2 \sqrt{x})^{s-4} \nonumber \\ && \pm \frac{s(s-4)(s-5)}{3!} (2 \sqrt{x})^{s-6} + \cdots \mbox{ to } [1+\half s] \mbox{ terms }, \end{eqnarray}
which is true for all positive integral values of $s$.

Similarly, we can shew that if $s$ is a positive even integer, then

\begin{eqnarray} \frac{s}{1!} 2^{s-2} && \{\zeta (1-\half s) - \zeta(1-\half s, x)\} \nonumber \\ &+& \frac{s(s-4)(s-5)}{3!} 2^{s-6} \{\zeta(3-\half s) - \zeta(3-\half s, x)\}\nonumber \\ &+& \frac{s(s-6)(s-7)(s-8)(s-9)}{5!} 2^{s-10} \{\zeta (5-\half s) - \zeta(5-\half s, x)\}\nonumber \\ &+& \cdots \mbox{ to } [\mbox{$\frac{1}{4}$} (s+2)] \mbox{ terms } \nonumber \\ &=& \half \{\sqrt{x} + \sqrt{(x-1)}\}^s + \half \{\sqrt{x} - \sqrt{(x-1)}\}^s - 1 . \end{eqnarray}
Now, remembering (7) and putting $x=1$ in (15), we obtain
\begin{eqnarray} \Phi (s, 0) &=& -\frac{s}{\sqrt{2}} \pi^{-\frac{1}{2}(1+s)} \cos \mbox{$\frac{1}{4}$} \pi s \{1\cdot 3\cdot 5 \cdots (s-2) \pi \zeta (\half s) \}\nonumber \\ \quad\quad &-& 3\cdot 5\cdot 7 \cdots (s-4) \half (s-5)\mbox{$\frac{1}{3}$} \pi^3 \zeta (\half s-2) \nonumber \\ \quad\quad &+& 5\cdot 7\cdot 9 \cdots (s-6) \half (s-7)\mbox{$\frac{1}{4}$} (s-9) \mbox{$\frac{1}{5}$} \pi^5 \zeta (\half s-4) \nonumber \\ \quad\quad &-& 7\cdot 9\cdot 11 \cdots (s-8) \half (s-9) \mbox{$\frac{1}{4}$} (s-11) \mbox{$\frac{1}{6}$} (s-13) \mbox{$\frac{1}{7}$} \pi^7 \zeta (\half s-6) \nonumber \\ \quad\quad &+& 9\cdot 11\cdot 13 \cdots (s-10) \half (s-11) \mbox{$\frac{1}{4}$} (s-13)\mbox{$\frac{1}{6}$}(s-15) \mbox{$\frac{1}{8}$} (s-17) \nonumber \\ && \quad\quad \times \mbox{$\frac{1}{9}$} \pi^9 \zeta (\half s -8) - \cdots \mbox{ to } [\mbox{$\frac{1}{4}$} (s+1)] \mbox{ terms } , \end{eqnarray}
If $s$ is an odd integer greater than 1. Similarly, putting $x=\frac{1}{2}$ in (15), we can express $\Phi (s, \frac{1}{2})$ in terms of $\zeta$-functions, if $s$ is an odd integer greater than 1.

4. It is also easy to shew that, if $$\Psi (s,x) = \sum^{n=\infty}_{n=0} \frac{\{\sqrt{(x+n)} + \sqrt{(x+n+1)}\}^{-s}}{\sqrt{\{(x+n)(x+n+1)\}}},$$ then

\begin{eqnarray} \Psi (s,x) &-& \half \frac{\{\sqrt{x}+ \sqrt{(x-1)}\}^s - \{\sqrt{x} - \sqrt{(x-1)}\}^s}{\sqrt{\{x(x-1)\}}} \nonumber \\ &=& \frac{s-2}{1!} 2^{s-2} \zeta (2 - \half s, x) + \frac{(s-4)(s-5)(s-6)}{3!} 2^{s-6} \zeta (4 - \half s, x) \nonumber \\ &+& \frac{(s-6)(s-7)(s-8)(s-9)(s-10)}{5!} 2^{s-10} \zeta (6 -\half s, x) \nonumber\\ &+& \cdots \mbox{ to } [\mbox{$\frac{1}{4}$} (s+1)] \mbox{ terms }, \end{eqnarray}
provided that $s$ is a positive odd integer. For example
\begin{eqnarray} \left.\begin{array}{ll} \Psi (1,x) = \frac{1}{\sqrt{x}}, \\ \Psi (3,x) = 4 \sqrt{x} - \frac{1}{\sqrt{x}} + 2 \zeta (\half , x),\\ \Psi (5,x) = 16 x \sqrt{x}-12 \sqrt{x} + \frac{1}{\sqrt{x}} + 24 \zeta (-\half ,x), \end{array}\right\} \end{eqnarray}
and so on.