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
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
$$ \sum_{n=1}^k T_n \approx \ln(k)+ \gamma $$
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