I want to find out the result of the following summation if some (maybe big) positive integer $m$ is given. $$ \sum_{i=1}^{m}\binom{m}{i}\frac{(-1)^{i+1}}{2^i-1} $$ It doesn't seem much possible to get an easy result since the denominator is $2^i-1$ rather than $2^i$. I tried to use Wolfram Alpha to test some values of $m$'s, and the result was close to $\log_2m$, e.g., it took 8.335...... when $m=256$ and 5.355...... when $m=32$. This may not be coincidence and I wonder if there are any connections between them.
1 Answer
By expanding $(2^i-1)^{-1}=\sum_{p=1}^\infty 2^{-ip}$, and summing over $i$, we find $$S_m=\sum_{i=1}^{m}\binom{m}{i}\frac{(-1)^{i+1}}{2^i-1}=\sum _{p=1}^{\infty } \left(1-\left(1-2^{-p}\right)^m\right).$$
This sum was considered in this post and in a publication On the expectation of the maximum of IID geometric random variables. The latter derives on page 140 an accurate expression in terms of the harmonic number $H_m$, $$S_m=\frac{H_{m}}{\ln 2}-\frac{1}{2}+\epsilon_m,\;\;\text{with}\;\;|\epsilon_m|<0.0006,\;\;\text{for}\;\;m\geq 10.$$
In the plot I compare $S_m$ (gold) with $H_m/\ln 2 - 1/2$ (blue), the difference is indeed nearly invisible.
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$\begingroup$ I don't have access to the source material, but it would be good to clarify whether $\varepsilon_m\in o_m(1)$ or whether this is one of the 'fourier-esque' tiny sums that sometimes show up in sums similar to this, small but not actually tending to zero. $\endgroup$ Nov 16, 2021 at 17:41
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$\begingroup$ @StevenStadnicki --- the paper is on Researchgate (freely downloadable) --- I don't think $\epsilon_m$ vanishes in the limit $m\rightarrow\infty$. $\endgroup$ Nov 16, 2021 at 17:53