# Examine whether the series ∞∑ 1/(logn)^logn∑ is convergent.n=2

Examine whether the series $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{\left(\mathrm{log}n{\right)}^{\mathrm{log}n}}$ is convergent. $n=2$

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aprovard

Notice that, for all $n⇒2$
$\left(\mathrm{log}n{\right)}^{\mathrm{log}n}={e}^{\mathrm{log}\left(\left(\mathrm{log}n{\right)}^{{\mathrm{log}}^{n}}\right)}=\left({e}^{\mathrm{log}n}{\right)}^{\mathrm{log}\left(\mathrm{log}n\right)}={n}^{\mathrm{log}\left(\mathrm{log}n\right)}$
Also note that $\mathrm{log}\left(\mathrm{log}n\right)>2$ for all $n>{e}^{{e}^{2}}$. Choose n0 such that ${n}_{0}>{e}^{{e}^{2}}$. Then for all $n⇒n0$ we have $\frac{1}{{\left(\mathrm{log}n\right)}^{\mathrm{log}n}}=\frac{1}{{n}^{\mathrm{log}}}\left(\mathrm{log}n\right)\le \frac{1}{{n}^{2}}$
The given series is convergent by comparison test.