For a real number $x$,

  • Floor Function returns the largest integer less than or equal to $x$, denote as $\lfloor x \rfloor$.

  • Ceiling Function returns the smallest integer larger than or equal to $x$, denote as $\lceil x \rceil$.

  • Sawtooth Function returns the fractional part of $x$, denote as $\{x\}$.

Help me to find the smallest positive real number $C$, such that for any positive integer $k$, $$S=\sum_{i=k-\lfloor \frac{k-1}{2} \rfloor+1}^k \dfrac{1}{i} +\sum_{i=k-\lceil \frac{k-1}{2} \rceil+1}^k \dfrac{1}{i} < C$$

Note that calculating for some $k$:

$k=11$-re $S=15796/13860\thickapprox 1.139754$.

$k=10001$-re $S\thickapprox 1.138626$.

$k=10^{10001}+1$-re $S\thickapprox 1.38629$.

So it is easy to find a claim.


1 Answer 1


It should be $C= \ln 4$. That's beacause for large $k$ the two sums are well approximated by $$\int_{k/2}^k \frac{1}{x} \ \mathrm{d}x +\int_{k/2}^k \frac{1}{x} \ \mathrm{d}x = \ln 2 + \ln 2 =2 \ln 2=\ln 4 \thickapprox 1.38629436...$$


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