A sum involving binomial coefficients and powers of 2 I am interested in a simplified version of the following sum 
$$\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^k}{2^k-1}.$$
I have to evaluate it for values of n ranging from $10^{4}$ till $10^{10}.$
Is there a way to express it in terms of some special function computable through Matlab or mathematica?
UPDATE : For small values of $n$ I noticed that the value is quite close to $−log_2(n).$
 A: We can rewrite the sum using a geometric series, then apply the binomial theorem:
\begin{align*}
\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^k}{2^k-1}&=\sum_{k=1}^n (-1)^k {n\choose k} \sum_{m=1}^\infty \frac{1}{2^{mk}}\\
&=\sum_{m=1}^{\infty}\left[\left(1-\frac{1}{2^{m}}\right)^n-1\right].
\end{align*}
In a sense this made things worse, because we replaced the finite sum with an infinite one. On the other hand, the infinite series is nice because:


*

*the terms are all negative and increase monotonically to $0$, and

*the terms decay exponentially once $m$ is bigger than $\log_2(n)$.


For $n$ large, terms in the series are very close to $-1$ when $m$ is less than $\log_2 n$ and very close to $0$ when $m$ is greater than $\log_2 n$. With some care, this is enough to show that the sum never differs from $-\log_2(n)$ by more than $2$.
A: Using Newton's Binomial Theorem  we can obtain a lower bound
$$\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^k}{2^k-1} \approx \sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^k}{2^k}=-1+\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{2^k}=-1+(1-\frac{1}{2})^n=-1+\frac{1}{2^n}$$
