Divergence of $\sum _{n\ge 2}\frac{1}{{\mathrm{ln}}^{p}n}$ for $1<p\le \mathrm{\infty}$

elisegayezm
2022-10-06
Answered

Divergence of $\sum _{n\ge 2}\frac{1}{{\mathrm{ln}}^{p}n}$ for $1<p\le \mathrm{\infty}$

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Ufumanaxi

Answered 2022-10-07
Author has **5** answers

By the integral test for convergence, you may write, for all $N\ge 3$,

$${\int}_{3}^{N}\frac{1}{{\mathrm{ln}}^{p}x}dx\le \sum _{3\le n\le N}\frac{1}{{\mathrm{ln}}^{p}n}$$

but, as $N\to \mathrm{\infty}$

$${\int}_{3}^{N}\frac{1}{{\mathrm{ln}}^{p}x}dx={\int}_{\mathrm{ln}3}^{\mathrm{ln}N}\frac{{e}^{t}}{{t}^{p}}dt\ge {\int}_{1}^{\mathrm{ln}N}{e}^{t}dt=N\phantom{\rule{thinmathspace}{0ex}}\to +\mathrm{\infty}$$

showing the divergence of the series $\sum _{n\ge 2}\frac{1}{{\mathrm{ln}}^{p}n}$, $1<p<\mathrm{\infty}$

$${\int}_{3}^{N}\frac{1}{{\mathrm{ln}}^{p}x}dx\le \sum _{3\le n\le N}\frac{1}{{\mathrm{ln}}^{p}n}$$

but, as $N\to \mathrm{\infty}$

$${\int}_{3}^{N}\frac{1}{{\mathrm{ln}}^{p}x}dx={\int}_{\mathrm{ln}3}^{\mathrm{ln}N}\frac{{e}^{t}}{{t}^{p}}dt\ge {\int}_{1}^{\mathrm{ln}N}{e}^{t}dt=N\phantom{\rule{thinmathspace}{0ex}}\to +\mathrm{\infty}$$

showing the divergence of the series $\sum _{n\ge 2}\frac{1}{{\mathrm{ln}}^{p}n}$, $1<p<\mathrm{\infty}$

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