sedeln5w

Answered

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.

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.

Answer & Explanation

kuncwadi17

Expert

2022-06-28Added 16 answers

$\begin{array}{rl}\sum _{i=0}^{N-1}\frac{1}{2i+1}& =\frac{1}{2}\sum _{i=0}^{N-1}\frac{1}{i+1/2}=\frac{1}{2}[\mathrm{\Psi}(\frac{1}{2}+N)-\mathrm{\Psi}\left(\frac{1}{2}\right)]=\frac{1}{2}\mathrm{\Psi}(N+\frac{1}{2})+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\gamma +\mathrm{ln}\left(2\right)\end{array}$

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, $\mathrm{\Psi}\left(z\right)\approx \mathrm{ln}\left(z\right)-\frac{1}{2z}\text{}{\textstyle \text{when}}\text{}\left|\phantom{\rule{thinmathspace}{0ex}}z\phantom{\rule{thinmathspace}{0ex}}\right|\gg 1$

$\begin{array}{rl}\sum _{i=0}^{N-1}\frac{1}{2i+1}& \approx \frac{1}{2}[\mathrm{ln}(N+\frac{1}{2})-\frac{1}{2(N+1/2)}]+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\gamma +\mathrm{ln}\left(2\right)\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{2em}{0ex}}N\gg 1\end{array}$

${\sum _{i=0}^{N-1}\frac{1}{2i+1}\approx \frac{1}{2}\mathrm{ln}(N+\frac{1}{2})-\frac{1}{2(2N+1)}+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\gamma +\mathrm{ln}\left(2\right)\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}N\gg 1}$

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, $\mathrm{\Psi}\left(z\right)\approx \mathrm{ln}\left(z\right)-\frac{1}{2z}\text{}{\textstyle \text{when}}\text{}\left|\phantom{\rule{thinmathspace}{0ex}}z\phantom{\rule{thinmathspace}{0ex}}\right|\gg 1$

$\begin{array}{rl}\sum _{i=0}^{N-1}\frac{1}{2i+1}& \approx \frac{1}{2}[\mathrm{ln}(N+\frac{1}{2})-\frac{1}{2(N+1/2)}]+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\gamma +\mathrm{ln}\left(2\right)\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{2em}{0ex}}N\gg 1\end{array}$

${\sum _{i=0}^{N-1}\frac{1}{2i+1}\approx \frac{1}{2}\mathrm{ln}(N+\frac{1}{2})-\frac{1}{2(2N+1)}+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\gamma +\mathrm{ln}\left(2\right)\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}N\gg 1}$

polivijuye

Expert

2022-06-29Added 16 answers

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

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