gabolzm6d

2022-04-06

Fix a positive number $\alpha$ and consider the series

$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{(k+1){\left[\mathrm{ln}(k+1)\right]}^{\alpha}}$

libertydragonrbha

Beginner2022-04-07Added 15 answers

Consider the function $\frac{1}{(x+1){\left[\mathrm{ln}(x+1)\right]}^{\alpha}}$ for $\alpha >0$ . Then $\frac{1}{(k+1){\left[\mathrm{ln}(k+1)\right]}^{\alpha}}$ is positive, decreasing, continuous on $[2,\mathrm{\infty})$ . Therefore, by integral test, we have

$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{(k+1){\left[\mathrm{ln}(k+1)\right]}^{\alpha}}$

converges if and only if

${\int}_{1}^{\mathrm{\infty}}\frac{1}{(x+1){\left[\mathrm{ln}(x+1)\right]}^{\alpha}}dx$

converges. On the other hand, if$\alpha >1$ , then

$\int}_{1}^{\mathrm{\infty}}\frac{1}{(x+1){\left[\mathrm{ln}(x+1)\right]}^{\alpha}}dx=\frac{1}{1-\alpha}{{\mathrm{ln}(x+1)}^{1-\alpha}{\mid}^{\mathrm{\infty}}}_{1}<\mathrm{\infty$

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

$\frac{1}{(k+1){\left[\mathrm{ln}(k+1)\right]}^{\alpha}}\ge \frac{1}{k+1}$

and the harmonic series$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{k+1}$ diverges. By comparison test,

$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{(k+1){\left[\mathrm{ln}(k+1)\right]}^{\alpha}}$ diverges when $\alpha \le 0$

converges if and only if

converges. On the other hand, if

and it diverges if

Also, for

and the harmonic series

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