# I ran into this limit while evaluating an integral. <munder> <mo movablelimits="true" form="

Joel French 2022-07-03 Answered
I ran into this limit while evaluating an integral.
$\underset{N\to \mathrm{\infty }}{lim}\left(2\sqrt{N+1}\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}\sum _{n=1}^{N}\frac{1}{\sqrt{n}}\right)$
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jugf5
First of all we see that eta-function is defined (the series converges) for $s>0$ (Dirichlet convergence test). Next, we use the relation between eta-function and zeta-function for $s>1$
$\eta \left(s\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n-1}}{{n}^{s}}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{s}}-2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(2n{\right)}^{s}}=\left(1-{2}^{1-s}\right)\zeta \left(s\right)$
Using this relation we can define zeta-function for $s\in \left(0;1\right)$ as $\zeta \left(s\right)=\frac{\eta \left(s\right)}{1-{2}^{1-s}}$ We can also present eta-function in the form
$\underset{N\to \mathrm{\infty }}{lim}\sum _{n=1}^{2N}\frac{\left(-1{\right)}^{n-1}}{{n}^{s}}=\underset{N\to \mathrm{\infty }}{lim}\left(\sum _{n=1}^{2N}\frac{1}{{n}^{s}}-2\sum _{n=1}^{N}\frac{1}{\left(2n{\right)}^{s}}\right)=\underset{N\to \mathrm{\infty }}{lim}\left(\sum _{n=1}^{N}\frac{1-{2}^{1-s}}{{n}^{s}}+\sum _{n=N+1}^{2N}\frac{1}{{n}^{s}}\right)$
The second term can be presented as the Riemann sum at $N\to \mathrm{\infty }$
${I}_{2}\left(N\right)=\sum _{n=N+1}^{2N}\frac{1}{{n}^{s}}=\frac{N}{{N}^{S}}\frac{1}{N}\sum _{n=N+1}^{2N}\frac{1}{\left(\frac{n}{N}{\right)}^{s}}\to {N}^{1-s}{\int }_{1}^{2}\frac{dx}{{x}^{s}}={N}^{1-s}\frac{{2}^{1-s}-1}{1-s}$
$\eta \left(s\right)=\underset{N\to \mathrm{\infty }}{lim}\left(\sum _{n=1}^{N}\frac{1-{2}^{1-s}}{{n}^{s}}-{N}^{1-s}\frac{1-{2}^{1-s}}{1-s}\right)$
$\zeta \left(s\right)=\underset{N\to \mathrm{\infty }}{lim}\left(\sum _{n=1}^{N}\frac{1}{{n}^{s}}-\frac{{N}^{1-s}}{1-s}\right)\phantom{\rule{thinmathspace}{0ex}};\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}s\in \left(0;1\right)$
Our initial limit
$\underset{N\to \mathrm{\infty }}{lim}\left(2\sqrt{N+1}\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}\sum _{n=1}^{N}\frac{1}{\sqrt{n}}\right)=\underset{N\to \mathrm{\infty }}{lim}\left(2\sqrt{N}\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}\sum _{n=1}^{N}\frac{1}{\sqrt{n}}\right)=-\zeta \left(\frac{1}{2}\right)=1.46035...$