Approximation of Sum Containing a Product

Question

Given the sum, $$\sum_{n=1}^k\sum_{j=0}^\infty\prod_{r=0}^{j}\frac{1}{n+rk}$$ I am trying to find an integral approximation, but any other approximation method would also be a good start. The obviously tricky part about this sum is the product it contains, but alas using the Stirling factorial approximation technique (turning the product into a sum with logarithms) has led me nowhere. Any help would be greatly appreciated.

Answer 1

Term $n$ is

$$T_n = \sum_{j=0}^\infty \prod_{r=0}^j \frac{1}{n+rk} = k^{n/k-1} e^{1/k} \left( \Gamma\left(\frac{n}{k}\right) - \Gamma\left(\frac{n}{k},\frac{1}{k}\right)\right)$$ If you want an integral, you can represent this as $$ T_n = \frac{e^{1/k}}{k} \int_0^{1} e^{-t/k} t^{n/k-1}\; dt $$ For large $k$, you can approximate $e^{(1-t)/k} \approx 1 + (1-t)/k$ making this $$ T_n \approx \frac{k+n+1}{n(k+n)}$$ and then your expression $$ \sum_{n=1}^k T_n \approx \ln(k)+ \gamma $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\sum_{n = 1}^{k}\sum_{j = 0}^{\infty}\prod_{r = 0}^{j}{1 \over n + rk} = \sum_{n = 1}^{k}{1 \over n}\sum_{j = 0}^{\infty} {1 \over \prod_{r = 1}^{j}\pars{r + n/k}} = \sum_{n = 1}^{k}{1 \over n}\sum_{j = 0}^{\infty} {1 \over \pars{1 + n/k}^{\,\overline{\,j\,}}} \\[5mm] = &\ \sum_{n = 1}^{k}{1 \over n}\sum_{j = 0}^{\infty} {1 \over \Gamma\pars{1 + n/k + j}/\Gamma\pars{1 + n/k}} \\[5mm] = &\ \sum_{n = 1}^{k}{1 \over n}\bracks{1 + \sum_{j = 1}^{\infty}{1 \over \pars{j - 1}!} {\Gamma\pars{1 + n/k}\Gamma\pars{j} \over \Gamma\pars{1 + n/k + j}}} = H_{k} + \sum_{n = 1}^{k}{1 \over n} \sum_{j = 0}^{\infty}{1 \over j!}\int_{0}^{1}t^{n/k}\pars{1 - t}^{\,j}\,\dd t \\[5mm] = &\ H_{k} + \int_{0}^{1}\exp\pars{1 - t} \sum_{n = 1}^{k}{t^{n/k} \over n}\,\dd t \\[5mm] & = \bbx{\ds{H_{k} - \int_{0}^{1}\exp\pars{1 - t}\bracks{% \ln\pars{1 - t^{1/k}} - \Phi\pars{t^{1/k},1,k + 1}t^{1 + 1/k}}\dd t}} \end{align}

$\ds{\Phi}$ is the Lerch Trascendent Function. I suspect there is not a 'closed form' for the last expression.