TITLE: Adding up a series
QUESTION [5 upvotes]: Is there any way to add that up in a reasonable fashion: 
$$
\frac{1}{4^1+1}+\frac1{4^2+2}+\cdots+\frac1{4^n+n}?
$$
It isn't geometric or arithmetic so I have no idea how to go about it besides just eye-balling it.

REPLY [5 votes]: That sum is quite easy to estimate since it is a little less than:
$$ \frac{1}{4}+\frac{1}{4^2}+\ldots+\frac{1}{4^n} = \frac{1}{3}\left(1-\frac{1}{4^n}\right) $$
but a little greater than:
$$ \frac{1}{4}\left(1-\frac{1}{4}\right)+\frac{1}{4^2}\left(1-\frac{2}{4^2}\right)+\ldots+\frac{1}{4^n}\left(1-\frac{n}{4^n}\right) = \frac{1}{225}\left(59-\frac{75}{4^n}+\frac{15n+16}{16^n}\right).$$
I see no reason for expecting a nice closed form for the finite sum.
About the series:
$$ \begin{eqnarray*}\sum_{k\geq 1}\frac{1}{4^k+k} = \sum_{k\geq 1}\sum_{m\geq 0}\frac{(-1)^{m} k^{m}}{4^{k(m+1)}}&=&\sum_{m\geq 0}\left.\frac{d^m}{dx^m}\sum_{k\geq 1}\frac{1}{4^k}\exp\left(-\frac{kx}{4^k}\right)\right|_{x=0}\\&=&\int_{0}^{+\infty}e^{-x}\sum_{k\geq 1}\frac{1}{4^k}\exp\left(-\frac{kx}{4^k}\right)\,dx\end{eqnarray*} $$
is close to $0.275624$ and can be written as 
$$\sum_{m\geq 0}(-1)^m\cdot\text{Li}_{-m}\left(\frac{1}{4^{m+1}}\right), $$
but that does not look promising.