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When $K$ is a positive integer and $1\le K \le 30$, how to prove the following expression:
$$ \sum_{t=1}^K \binom{K}{t} (-1)^t \log_2 t \le \log_2 K $$

Through simulation, I found that the expression is indeed correct, but I don't know how to prove it.

I tried to discuss the odd and even cases $K$ separately. I'm not sure if that's the right idea.

PS: The following link has the approximation for $ \sum_{t=1}^K \binom{K}{t} (-1)^t \log_2 t $, but I still can not prove the problem.

Proof $\sum\limits_{k=1}^n \binom{n}{k}(-1)^k \log k = \log \log n + \gamma +\frac{\gamma}{\log n} +O\left(\frac1{\log^2 n}\right)$

fredal
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1 Answers1

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Using Frullani integral $$\log a=\int_0^\infty\frac{e^{-x}-e^{-ax}}{x}\,dx=\int_0^1\frac{1-t^{a-1}}{-\log t}\,dt,\qquad(a>0)$$ we get an exact formula $$s_n:=\sum_{k=1}^n\binom{n}{k}(-1)^k\log k=\int_0^1\frac{1-t-(1-t)^n}{-t\log t}\,dt=\int_0^1\frac{1-x^{n-1}}{1-x}\frac{-x}{\log(1-x)}\,dx.$$ Hence, if we put $f_n(x)=(1-x^{n-1})/(1-x)$ and $g(x)=x/\log(1-x)$, we get $$\log n-s_n=\int_0^1 f_n(x)\big(g(x)-g(1-x)\big)\,dx,$$ and if we split $\int_0^1=\int_0^{1/2}+\int_{1/2}^1$ and substitute $x=1-t$ in the last integral, we obtain $$\log n-s_n=\int_0^{1/2}\big(f_n(1-t)-f_n(t)\big)\big(g(1-t)-g(t)\big)dt\geqslant 0$$ since both $f_n$ and $g$ are increasing (which is easy to see from the power series of $f_n$ and $1/g$).

metamorphy
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  • (Perhaps there's also an elementary solution using induction, but I didn't find one.) – metamorphy Aug 29 '20 at 10:28
  • Thank you. The answer is very professional and helpful. But I have a new question: How to prove $f_n$ is increasing for $0<x<1/2$. The power series is ${{f}{n}}\left( x \right)=\sum\limits{k=0}^{\infty }{\left( \begin{matrix} n-1 \ k+1 \ \end{matrix} \right){{\left( x-1 \right)}^{k}}}$. when $x-1<0$, I cannot sure the result is increasing. – fredal Aug 30 '20 at 02:55
  • @fredal: $f_n(x)=\sum_{k=0}^{n-2}x^k$ if $n>1$. – metamorphy Aug 30 '20 at 05:33
  • Got it. Thank you very much. – fredal Aug 30 '20 at 06:52