sedeln5w

2022-06-27

Logarithmic approximation of $\sum _{0}^{N-1}\frac{1}{2i+1}$
Can anyone confirm that it's possible to approximate the sum $\sum _{0}^{N-1}\frac{1}{2i+1}$ with the $\frac{\mathrm{log}N}{2}$? And why?
It's clearly visible that the sum has a logarithmic growth over i but it's unclear to me how to prove it.

Expert

$\begin{array}{rl}\sum _{i=0}^{N-1}\frac{1}{2i+1}& =\frac{1}{2}\sum _{i=0}^{N-1}\frac{1}{i+1/2}=\frac{1}{2}\left[\mathrm{\Psi }\left(\frac{1}{2}+N\right)-\mathrm{\Psi }\left(\frac{1}{2}\right)\right]=\frac{1}{2}\mathrm{\Psi }\left(N+\frac{1}{2}\right)+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\gamma +\mathrm{ln}\left(2\right)\end{array}$
where $\mathrm{\Psi }\left(z\right)$ is the $\mathit{\text{digamma function}}$ and $\gamma =0.577215664901533\dots$ is the $\mathit{\text{Euler-Mascheroni constant}}$. $\mathrm{\Psi }\left(\frac{1}{2}\right)=-\gamma -2\mathrm{ln}\left(2\right)$
Also,
$\begin{array}{rl}\sum _{i=0}^{N-1}\frac{1}{2i+1}& \approx \frac{1}{2}\left[\mathrm{ln}\left(N+\frac{1}{2}\right)-\frac{1}{2\left(N+1/2\right)}\right]+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\gamma +\mathrm{ln}\left(2\right)\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{2em}{0ex}}N\gg 1\end{array}$
$\sum _{i=0}^{N-1}\frac{1}{2i+1}\approx \frac{1}{2}\mathrm{ln}\left(N+\frac{1}{2}\right)-\frac{1}{2\left(2N+1\right)}+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\gamma +\mathrm{ln}\left(2\right)\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}N\gg 1$

polivijuye

Expert

This is a Riemann sum for the integral from 0 to $N$ of $1/\left(2x+1\right)$, so it approximates $\left(1/2\right)\mathrm{ln}\left(2N+1\right)$, which is approximately $\left(1/2\right)\mathrm{ln}N$