1. Prof. Sanjana remarks that it is not easy to evaluate the
series
$$\frac{1}{1^n} + \frac{1}{2} \frac{1}{3^n} + \frac{1\cdot 3}{2\cdot 4}
\frac{1}{5^n} + \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \frac{1}{7^n} + \cdots
\mbox{ ad inf.,}$$
if $n \gt 3$. In attempting to sum the series for all values of $n$,
I have arrived at the following results:
Let
\begin{eqnarray*}
f(p) &=& \frac{1}{1+p} + \frac{1}{2} \frac{1}{3+p} +
\frac{1\cdot 3}{2\cdot 4} \frac{1}{5+p} + \cdots \\
&=& \int\limits^1_0 x^p \left(1 + \frac{1}{2} x^2 +
\frac{1\cdot 3}{2\cdot 4} x^4 + \cdots \right) dx \\
&=& \int\limits^1_0 \frac{x^p}{\sqrt{1-x^2}} dx = \myfrac{1}{2}
\int\limits^1_0 x^{\frac{1}{2}(p-1)} (1-x)^{- \frac{1}{2}} dx\\
&=& \myfrac{1}{2} \frac{\Gamma \left(\frac{p+1}{2}\right) \Gamma
(\frac{1}{2})}
{\Gamma \left(\frac{p+2}{2}\right)} = \frac{\pi^{\frac{1}{2}}}{2}
\frac{\Gamma\left(\frac{p+1}{2}\right)}{\Gamma
\left(\frac{p+2}{2}\right)}.
\end{eqnarray*}
But
$$\Gamma \left(\frac{p+1}{2}\right) =
\ds\frac{\pi^{\frac{1}{2}}}{2^p}
\frac{\Gamma(p+1)}{\Gamma
\left(\frac{p+2}{2}\right)}
$$
(vide Williamson, Integral Calculus, p. 164).
Therefore
$$\ds f(p) = \frac{\pi}{2^{p+1}}
\frac{\Gamma(p+1)}{\left\{\Gamma \left(\frac{p+2}{2}\right)\right\}^2}.$$
Therefore
Again, by expanding $f(p)$ in ascending powers of $p$, we have \begin{eqnarray*} f(p) &=& \left(1+\frac{1}{2}\frac{1}{3} + \frac{1\cdot 3}{2\cdot 4} \frac{1}{5} + \cdots \right) - p \left(1+\frac{1}{2} \frac{1}{3^2} + \frac{1\cdot 3}{2\cdot 4} \frac{1}{5^2} + \cdots \right) + p^2 \left(1 + \frac{1}{2} \frac{1}{3^3} + \frac{1\cdot 3}{2\cdot 4} \frac{1}{5^3} + \cdots \right) - \cdots\\ &=& \frac{\pi}{2} \{\phi (0) - p \phi (1) + p^2 \phi (2) - p^3 \phi (3) + \cdots \}, \end{eqnarray*} where $$\frac{1}{1^{n+1}} + \frac{1}{2} \frac{1}{3^{n+1}} + \frac{1\cdot 3}{2\cdot 4} \frac{1}{5^{n+1}} + \cdots \equiv \frac{\pi}{2} \phi (n).$$ Hence (1) may be written \begin{eqnarray*} \log \myfrac{1}{2}\pi + \log \{\phi (0) - p \cdot \phi(1) + p^2 \cdot \phi (2) - \cdots \} &=& \log (\myfrac{1}{2} \pi) - p \log 2 + \frac{p^2}{2} \left(1 - \frac{1}{2} \right) S_2 - \frac{p^3}{3} \left(1 - \frac{1}{2^2} \right) S_3 + \cdots \\ &=& \log (\myfrac{1}{2} \pi) - p \sigma _1 + \frac{p^2}{2} \sigma_2 - \frac{p^3}{3} \sigma_3 + \cdots, \end{eqnarray*} where $$ \sigma_n \equiv 1 - \frac{1}{2^n} + \frac{1}{3^n} - \frac{1}{4^n} + \cdots.$$ Differentiating with respect to $p$, and equating the coefficients of $p^{n-1},$ we have $$n \phi (n) \equiv \sigma_1 \phi (n-1) + \sigma_2 \phi (n-2) + \sigma_3 \phi (n-3) + \cdots \mbox{ to } n \mbox{ terms}.$$ Thus we see that \begin{eqnarray*} \frac{\pi}{2} \phi(0) \equiv 1 + \frac{1}{2}\frac{1}{3}{\phantom{1}} + \frac{1\cdot 3}{2\cdot 4} \frac{1}{5}{\phantom{1}} + \cdots &=& \frac{\pi}{2}, \\ \frac{\pi}{2}\phi (1) \equiv 1 + \frac{1}{2} \frac{1}{3^2} + \frac{1\cdot 3}{2\cdot 4}\frac{1}{5^2} +\cdots &=& \frac{\pi}{2} (\log 2), \\ \frac{\pi}{2} \phi(2) \equiv 1 + \frac{1}{2} \frac{1}{3^3} + \frac{1\cdot 3}{2\cdot 4} \frac{1}{5^3} + \cdots &=& \frac{\pi^3}{48}+ \frac{\pi}{4} (\log 2)^2, \\ \frac{\pi}{2} \phi (3) \equiv 1 +\frac{1}{2} \frac{1}{3^4} + \frac{1\cdot 3}{2\cdot 4} \frac{1}{5^4} + \cdots &=& \frac{\pi^3}{48} \log 2 + \frac{\pi^3}{12} (\log 2)^3 + \frac{\pi}{6} \sigma_3\\ &=&\frac{\pi^3}{48} \log 2 + \frac{\pi ^3}{12} (\log 2)^3 + \frac{\pi}{8} S_3, \end{eqnarray*} and so on.
2. More generally, consider the series $$ \frac{1}{b^n} -\frac{a}{1!} \frac{1}{(b+1)^n} + \frac{a(a-1)}{2!} \frac{1}{(b+2)^n} - \cdots .$$ Writing $$\frac{\Gamma (b) \Gamma(a+1)}{\Gamma(a+b+1)} \phi(n-1)$$ for this, and taking the identity $$ \frac{1}{b+p} - \frac{a}{1!} \frac{1}{b+1+p} + \frac{a(a-1)}{2!} \frac{1}{b+2+p} - \cdots =\int\limits^1_0 x^{b+p-1} (1-x)^a dx = \frac{\Gamma(b+p) \Gamma(a+1)}{\Gamma (a+b+p+1)} \ , $$ we find $$ n \phi (n) = \sigma_1 \phi (n-1) + \sigma_2 \phi (n-2) + \sigma_3 \phi (n-3) + \cdots \mbox{ to } n \mbox{ terms},$$ where $$ \sigma_n \equiv \frac{1}{b^n} - \frac{1}{(a+b+1)^n} + \frac{1}{(b+1)^n} - \frac{1}{(a+b+2)^n} + \cdots \ .$$
Examples: Put $ a = -\frac{1}{2}, b= \frac{1}{4}.$ Then we see that \begin{eqnarray*} &(i)& 1 + \frac{1}{2}\frac{1}{5}{\phantom{1}} + \frac{1\cdot 3}{2\cdot 4} \frac{1}{9}{\phantom{1}} + \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{13} + \cdots = \frac{\left\{\Gamma(\frac{1}{4})\right\}^2}{4 \sqrt{(2 \pi)}},\\ &(ii)& 1+\frac{1}{2} \frac{1}{5^2} + \frac{1\cdot 3}{2\cdot 4} \frac{1}{9^2} + \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{13^2} + \cdots = \frac{\left\{\Gamma (\frac{1}{4})\right\}^2}{4 \sqrt{(2 \pi)}}\frac{\pi }{4}, \\ &(iii)& 1 + \frac{1}{2}\frac{1}{5^3} + \frac{1\cdot 3}{2\cdot 4}\frac{1}{9^3} + \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \frac{1}{13^3} + \cdots = \frac{\left\{\Gamma(\frac{1}{4})\right\}^2}{4 \sqrt{(2 \pi)}} \left\{\frac{\pi^2}{32}+\frac{1}{2}S'_2 \right\},\\ &(iv)& 1+ \frac{1}{2}\frac{1}{5^4} +\frac{1\cdot 3}{2\cdot 4}\frac{1}{9^4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \frac{1}{13^4} + \cdots = \frac{\left\{\Gamma(\frac{1}{4})\right\}^2}{4\sqrt{(2 \pi)}} \left\{\frac{5 \pi^3}{384} + \frac{\pi}{8} S'_2 + \frac{1}{3} S'_3\right\}, \end{eqnarray*} where $$S'_r = \frac{1}{1^r} - \frac{1}{3^r} + \frac{1}{5^r} - \frac{1}{7^r} + \cdots.$$