$a_n=\sum _{n=1}^k\left(\frac{1}{\left(k-n+1\right) n}\right)$
This looks quite simple but I'm not able to write this in a closed form. I encountered this while trying to analyse the coefficients in $\sum_{n=2}^{\infty}a_n t^n$ for its growth, where $t$ is a variable. Perhaps the closed form would help to analyse things better. I also need to check whether there exists any relation among the coefficients.
P.S: Absolute convergence can be taken care.
Thank you in advance
(I swapped $n$ and $k$.)
$\begin{array}\\ a_n &=\sum _{k=1}^n\left(\dfrac{1}{(n-k+1) k}\right)\\ &=\dfrac1{n+1}\sum _{k=1}^n\left(\dfrac1{k}+\dfrac1{n-k+1}\right)\\ &=\dfrac1{n+1}\left(\sum _{k=1}^n\dfrac1{k}+\sum _{k=1}^n\dfrac1{n-k+1}\right)\\ &=\dfrac1{n+1}\left(\sum _{k=1}^n\dfrac1{k}+\sum _{k=1}^n\dfrac1{k}\right)\\ &=\dfrac{2}{n+1}\sum _{k=1}^n\dfrac1{k}\\ &=\dfrac{2}{n+1}(\ln(n)+\gamma+o(1))\\ \end{array} $