Help me with this sum: $\sum_{i=0}^\infty \frac{n}{\log^2\left(\frac{n}{5^i}\right)}$

Question

I have to prove something but I'm stuck, I ended up with this sum. Is there any transformation I can do in the following sum? $$\sum_{i=0}^\infty \frac{n}{\log^2\left(\frac{n}{5^i}\right)}$$

Answer 1

$\displaystyle\sum_{k=0}^\infty\frac1{\ln\bigg(\dfrac n{a^k}\bigg)^2}=\frac{\psi^{(1)}\big(-\log_an\big)}{\ln^2a}$ , see polygamma function for more details. For $n=1$, if the sum starts at $k=1$, the result is $\dfrac{\zeta(2)}{\ln^2a}=\dfrac{\pi^2}{6\ln^2a}$ , see Basel problem. Both expressions are derived by using some basic properties of the logarithmic function, such as $\ln\dfrac ab=\ln a-\ln b$, and $\ln a^b=$ $=b\ln a$, etc.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ A $\ul{detailed}$ derivation is given below:

\begin{align} \color{#00f}{\large\sum_{i = 0}^{\infty}{n \over \ln^{2}\pars{n/5^i}}}&= n\sum_{i = 0}^{\infty}{1 \over \bracks{\ln\pars{n} - \ln\pars{5}i}^{2}} ={n \over \ln^{2}\pars{5}} \sum_{i = 0}^{\infty}{1 \over \braces{\bracks{-\ln\pars{n}/\ln\pars{5}} + i}^{2}} \\[3mm]&=\color{#00f}{\large% {n \over \ln^{2}\pars{5}}\,\Psi'\pars{-\,{\ln\pars{n} \over \ln\pars{5}}}}\,, \qquad{-\,{\ln\pars{n} \over \ln\pars{5}}}\not=0,-1,-2,\ldots \end{align}

where $\ds{\Psi'\pars{z}}$ is the Trigamma Function. See ${\bf 6.4.10}$ in this table.