Prove absolute convergence from alternants

Question

He failed to get the show in its entirety in this series, one I could indicate how working with this kind of series? $$ \sum \limits^{\propto }_{n=1}\frac{(-1)}{n(\ln(n+1))^{2}} $$

Answer 1

We have $$\sum_{n\geq1}\left|\frac{-1}{n\log^{2}\left(n+1\right)}\right|=\sum_{n\geq1}\frac{1}{n\log^{2}\left(n+1\right)}=\frac{1}{\log^{2}\left(2\right)}+\sum_{n\geq2}\frac{1}{n\log^{2}\left(n+1\right)}\leq\frac{1}{\log^{2}\left(2\right)}+\sum_{n\geq2}\frac{1}{n\log^{2}\left(n\right)}\leq\frac{1}{\log^{2}\left(2\right)}+\frac{1}{2\log^{2}\left(2\right)}+\int_{2}^{\infty}\frac{1}{t\log^{2}\left(t\right)}dt $$ by integral test. And so $$\sum_{n\geq1}\left|\frac{-1}{n\log^{2}\left(n+1\right)}\right|\leq\frac{1}{\log^{2}\left(2\right)}+\frac{1}{2\log^{2}\left(2\right)}+\frac{1}{\log\left(2\right)}. $$ Note that if the series is $$\sum_{n\geq1}\frac{\left(-1\right)^{n}}{n\log^{2}\left(n+1\right)} $$ you can conclude that it is convergent by alternating series test.

Answer 2

We may use Cauchy condensation test. $$\sum_{n=1}^\infty{1\over{n(\ln(n+1))^2}}=\sum_{n=2}^\infty{1\over{(n-1)(\ln(n))^2}}<\infty\\ \iff \sum_{n=2}^\infty{e^n\over{(e^n-1)(\ln(e^n))^2}}<\infty\\ \iff\sum_{n=1}^\infty{e^n\over{(e^n-1)n^2}}<\infty\\ \iff\sum_{n=1}^\infty{1\over{n^2}}+\sum_{n=1}^\infty{1\over{(e^n-1)n^2}}<\infty $$ $${1\over (e^n-1)n^2}<{1\over n^2}$$ So by comparison test we see that the given series converges.