gabolzm6d

2022-04-06

Fix a positive number $\alpha$ and consider the series
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\left(k+1\right){\left[\mathrm{ln}\left(k+1\right)\right]}^{\alpha }}$

libertydragonrbha

Consider the function $\frac{1}{\left(x+1\right){\left[\mathrm{ln}\left(x+1\right)\right]}^{\alpha }}$ for $\alpha >0$. Then $\frac{1}{\left(k+1\right){\left[\mathrm{ln}\left(k+1\right)\right]}^{\alpha }}$ is positive, decreasing, continuous on $\left[2,\mathrm{\infty }\right)$. Therefore, by integral test, we have
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\left(k+1\right){\left[\mathrm{ln}\left(k+1\right)\right]}^{\alpha }}$
converges if and only if
${\int }_{1}^{\mathrm{\infty }}\frac{1}{\left(x+1\right){\left[\mathrm{ln}\left(x+1\right)\right]}^{\alpha }}dx$
converges. On the other hand, if $\alpha >1$, then
${\int }_{1}^{\mathrm{\infty }}\frac{1}{\left(x+1\right){\left[\mathrm{ln}\left(x+1\right)\right]}^{\alpha }}dx=\frac{1}{1-\alpha }{{\mathrm{ln}\left(x+1\right)}^{1-\alpha }{\mid }^{\mathrm{\infty }}}_{1}<\mathrm{\infty }$
and it diverges if $0<\alpha \le 1$. Therefore, the series converges when $\alpha >1$ and diverges when $0<\alpha <1$
Also, for $\alpha \le 0$, we have
$\frac{1}{\left(k+1\right){\left[\mathrm{ln}\left(k+1\right)\right]}^{\alpha }}\ge \frac{1}{k+1}$
and the harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k+1}$ diverges. By comparison test,
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\left(k+1\right){\left[\mathrm{ln}\left(k+1\right)\right]}^{\alpha }}$ diverges when $\alpha \le 0$

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