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
\begin{equation}
\zeta (s, 1) = \zeta (s), \zeta (s, \half ) = (2^s - 1) \zeta (s),
\end{equation}
where $\zeta(s)$ is the Riemann $\zeta$-function;
\begin{equation}
\zeta (s, x) - \zeta(s, x + 1) = x^{-s};
\end{equation}
\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,
\begin{equation}
6 \zeta (-\half , x) + (4x - 3) \sqrt{x} + \Phi (3,x) = 0.
\end{equation}
Similarly, we can shew that
\begin{equation}
40 \zeta (-\frac{3}{2}, x) + (16 x^2 - 20x + 5) \sqrt{x} + \Phi (5, x) =
0.
\end{equation}
2. Remembering the functional equation satisfied by $\zeta(s) $, viz.,
\begin{equation}
\zeta (1-s) = 2 (2 \pi)^{-s} \Gamma (s) \zeta(s) \cos \half \pi s,
\end{equation}
we see from (3) and (5) that
\begin{equation}
\sqrt{1} + \sqrt{2} + \sqrt{3} + \cdots + \sqrt{n} = \frac{2}{3}
n^{\frac{3}{2}} + \half \sqrt{n} - \frac{1}{4 \pi} \zeta (\frac{3}{2}) +
\frac{1}{6} \Phi (3, n);
\end{equation}
and
\begin{eqnarray}
\sqrt{1} &+& \sqrt{3} + \sqrt{5} + \cdots + \sqrt{(2n-1)} \nonumber \\
&=& \frac{1}{3} (2n - 1)^{\frac{3}{2}} + \half \sqrt{(2n-1)} +
\frac{\sqrt{2-1}}{4 \pi} \zeta (\frac{3}{2}) + \frac{1}{3\sqrt{2}} \Phi (3, n
- \half ).
\end{eqnarray}
Similarly from (6), we have
\begin{eqnarray}
1 \sqrt{1} &+& 2 \sqrt{2} + 3 \sqrt{3} + \cdots + n\sqrt{n} \nonumber\\
&=& \frac{2}{5} n^{\frac{5}{2}} + \half n^{\frac{3}{2}} + \frac{1}{8}
\sqrt{n} - \frac{3}{16 \pi^2} \zeta(\frac{5}{2}) + \frac{1}{40} \Phi (5, n);
\end{eqnarray}
and
\begin{eqnarray}
1 \sqrt{1} &+& 3 \sqrt{3} + 5 \sqrt{5} + \cdots + (2n-1)
\sqrt{(2n-1)} \nonumber \\
&=& \frac{1}{5} (2n-1)^{\frac{5}{2}} + \half (2n-1)^{\frac{3}{2}} +
\frac{1}{4} \sqrt{(2n-1)} \nonumber \\
&+& \frac{3(2\sqrt{2-1}}{16 \pi^2} \zeta (\mbox{$\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
\begin{equation}
\Phi (3,0) = \frac{3}{2 \pi} \zeta~(\frac{3}{2}), \Phi(5,0) =
\frac{15}{2 \pi^2} \zeta
(\frac{5}{2}).
\end{equation}
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.