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.
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Are you looking for a numerical value, or an exact answer in terms of other, well-known constants? – robjohn Oct 31 '15 at 23:52
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I wouldn't mind to know how to find an exact value, I think it is a great problem. It's not geometric or arithmetic – imranfat Nov 01 '15 at 00:43
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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.
Jack D'Aurizio
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