$\sum_{n=2}^\infty \frac{1}{(\ln\, n)^2}$ c0nvergence

Question

ps

La serie converge?

Verifique mi solución a continuación

Answer 1

Diverge Ver que$$\lim_{n\to\infty}\frac{\sqrt{n}}{\ln^2(n)}=\infty$$ Infact the power of $ n$ is $ \ frac {1} {2} \ leq1$ and $ \ lim_ {n \ to \ infty} n ^ {1/2} u_n = \ infty $

Answer 2

$\frac{1}{n}\leq\frac{1}{(ln\, n)^{2}}\iff n\geq(ln\, n)^{2}\iff ln\, n\leq\sqrt{n}\iff n\leq e^{\sqrt{n}}$

$ \begin{cases}(e^{\sqrt{n}})'>(n)'\\\text{for } n=1\,\,\,\, e^{1}>1\end {cases} $

Entonces, de la prueba de comparación, la serie diverge

Answer 3

Utilice la prueba de condensación de Cauchy

$$\sum{k=2}^{\infty} \frac{1}{(\ln k)^{2}} \text{converges} \iff \sum{k=2}^{\infty} \frac{2^k}{(\ln 2^k)^{2}} \text{converges}$$

$$\sum{k=2}^{\infty} \frac{2^k}{(\ln 2^k)^{2}} = \sum{k=2}^{\infty} \frac{2^k}{(\ln 2^k)(\ln 2^k)}=\sum{k=2}^{\infty} \frac{2^k}{k^2(\ln 2)(\ln 2)}=\sum{k=2}^{\infty} \frac{2^k}{k^2(\ln 2)^2} \text{(diverges)}$$

Así que diverge la serie original.