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Compute the value of the sum $\sum u_n$ where $u_{n}=\frac{(-1)^{n}}{n} \times\left\lfloor\frac{\ln (n)}{\ln (2)}\right\rfloor$.

I already know that this sum converges (see for example Show that the series $\sum_{n>0} \frac{(-1)^n}{n}\left \lfloor \frac{\ln n}{\ln 2} \right \rfloor$ converges ).

But I have no idea how to find the value of $\sum u_n$

MiKiDe
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    Is there a reason to think $\sum u_n$ has a simple closed form? You can get an arbitrarily close numerical approximation though, since the contributing factor is $\frac{\ln n}n$ and the map $n\mapsto\frac{\ln n}n$ is strictly decreasing for $n\ge 2$, so the terms get smaller and smaller pretty quickly in absolute value. – Prasun Biswas Oct 06 '21 at 21:37
  • The only thing I can think of: For $2^k\leq n<2^{k+1},$ $\lfloor \log n/\log 2\rfloor =k.$ $$\sum_{n=2^{k}}^{2^{k+1}-1} \frac{(-1)^n}{n}\left\lfloor \frac{\log n}{\log 2}\right\rfloor =k\sum_{n=2^k}^{2^{k+1}-1}\frac{(-1)^n}{n}.$$ But I don’t see how that helps. – Thomas Andrews Oct 06 '21 at 22:24

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We have $$\sum_{n=1}^\infty\frac{(-1)^n}{n}\lfloor\log_2(n)\rfloor = \gamma$$ the Euler–Mascheroni constant. See here for a proof. This was proved by G. Vacca in his article "A new series for the Eulerian constant γ = .577 ···", Quart. J. Pure Appl. Math. 41(1910), 363–364.

jjagmath
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