How can I get a closed form for $\sum_{k=1}^C \dfrac 1 k e^{-k} $?

Question

I am trying to get a closed form for

$$\sum_{k=1}^C \frac 1 k e^{-k} $$

Can anyone could help me?

Answer 1

Observe you have \begin{align} \sum^{N-1}_{k=0} x^k = \frac{1-x^{N}}{1-x} \ \ \Rightarrow \ \ \sum^{N-1}_{k=0}\frac{x^{k+1}}{k+1}=\sum^{N-1}_{k=0}\int^x_0 t^k\ dt =\int^x_0 \frac{1-t^{N}}{1-t}\ dt. \end{align} Re-indexing leads to \begin{align} \sum^N_{k=1} \frac{x^k}{k} =\int^x_0 \frac{1-t^{N}}{1-t}\ dt \end{align} which means \begin{align} \sum^N_{k=1} \frac{e^{-k}}{k} =\int^{e^{-1}}_0 \frac{1-t^N}{1-t}\ dt. \end{align}

Answer 2

You may write it as,

$$\sum_{k=1}^{\infty} \frac{(e^{-1})^k}{k}-\sum_{k=c+1}^{\infty} \frac{(e^{-1})^k}{k}$$

Then attempt to write it in terms of the Lerch transcendent $\Phi$. Using $\int_{0}^{1} x^{k-1} dx=\frac{1}{k}$,

$$=\sum_{k \geq 1} (e^{-1})(e^{-1})^{k-1} \int_{0}^{1} x^{k-1} dx-\sum_{k=0}^{\infty} \frac{(e^{-1})^{c+1+k}}{{c+1+k}}$$

Some algebra, and interchanging integral gives,

$$=\int_{0}^{1} \sum_{k \geq 1} (e^{-1})(e^{-1})^{k-1}x^{k-1} dx-e^{-c-1}\sum_{k=0}^{\infty} \frac{(e^{-1})^{k}}{{c+1+k}}$$

Recognizing the geometric series and $\Phi$, this is,

$$=e^{-1} \int_{0}^{1} \frac{1}{1-e^{-1}x} dx-e^{-c-1} \Phi (e^{-1},1,c+1)$$

Evaluating the integral gives,

$$=1-\log(e-1)-e^{-c-1} \Phi (e^{-1},1,c+1)$$

Answer 3

Mathematica confirms Ahmed S. Attaalla's answer:

$-e^{-c-1} \left(\Phi \left(\frac{1}{e},1,c+1\right)+e^{c+1} \log \left(\frac{e-1}{e}\right)\right)$