Lubna Boyce

2022-02-28

I am given the series
$\sum _{n=1}^{\mathrm{\infty }}\frac{{n}^{n-1}}{{\left(2{n}^{2}+n+1\right)}^{\left(n+\frac{1}{2}\right)}}$

shotokan0758s

The ratio test works fine
${a}_{n}=\frac{{n}^{n-1}}{{\left(2{n}^{2}+n+1\right)}^{\left(n+\frac{1}{2}\right)}}$
Take loragithms
$\mathrm{log}\left({a}_{n}\right)=\left(n-1\right)\mathrm{log}\left(n\right)-\left(n+\frac{1}{2}\right)\mathrm{log}\left(2{n}^{2}+n+1\right)$
Using Taylor for large values of n.
$\mathrm{log}\left({a}_{n}\right)=n\mathrm{log}\left(\frac{1}{2n}\right)+\frac{1}{2}\left(\mathrm{log}\left(\frac{1}{2{n}^{4}}\right)-1\right)-\frac{5}{8n}+O\left(\frac{1}{{n}^{2}}\right)$
Apply it twice and continue with Taylor series
$\mathrm{log}\left({a}_{n+1}\right)-\mathrm{log}\left({a}_{n}\right)=\left(\mathrm{log}\left(\frac{1}{2n}\right)-1\right)-\frac{5}{2n}$
$\frac{{a}_{n+1}}{{a}_{n}}={e}^{\mathrm{log}\left({a}_{n+1}\right)-\mathrm{log}\left({a}_{n}\right)}=\frac{1}{2en}\left(1-\frac{5}{2en}+O\left(\frac{1}{{n}^{2}}\right)\right)$
Now, using the expansion of $\mathrm{log}\left({a}_{n}\right)$ to $O\left(\frac{1}{n}\right)$, we then have your formula
${a}_{n}\sim \frac{1}{\sqrt{2e}}\frac{1}{{2}^{n}{n}^{n+2}}$
which is an overestimate of the true ${a}_{n}$

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