Proof-the-series-n-1-2-9-2n-ln-n-2-convergent-

Question Number 133568 by bemath last updated on 23/Feb/21

Proof the series \underset{n = 1}{\sum^{\infty}} \frac{2}{9 + 2 n {(\ln n)}^{2}}

convergent

Answered by EDWIN88 last updated on 23/Feb/21

let b_{n} = \frac{2}{9 + 2 n {(\ln n)}^{2}} and a_{n} = \frac{2}{2 n {(\ln n)}^{2}}

we know that \frac{2}{9 + 2 n {(\ln n)}^{2}} ⩽ \frac{2}{2 n {(\ln n)}^{2}} = \frac{1}{n {(\ln n)}^{2}}

then \underset{n = 1}{\sum^{\infty}} \frac{2}{9 + 2 n {(\ln n)}^{2}} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(\ln n)}^{2}}

consider \lim_{n \to \infty} \frac{1}{n {(\ln n)}^{2}} = 0, it follows that

\underset{n = 1}{\sum^{\infty}} \frac{1}{n {(\ln n)}^{2}} convergent, then \underset{n = 1}{\sum^{\infty}} \frac{2}{9 + 2 n {(\ln n)}^{2}}

also convergent

Answered by mathmax by abdo last updated on 24/Feb/21

\frac{2}{9 + 2 n {(lnn)}^{2}} ⩽ \frac{1}{n (\ln {(n)}^{2}} \Rightarrow \sum_{n = 1}^{\infty} \frac{2}{9 + 2 n {(lnn)}^{2}} ⩽ \sum_{n = 2}^{\infty} \frac{1}{n {(lnn)}^{2}}

the serie u_{n} = \frac{1}{n {(lnn)}^{2}} is decreazing to o so its nature is same to

\int_{2}^{\infty} \frac{dx}{x {(lnx)}^{2}} and changement lnx = t give

\int_{2}^{\infty} \frac{dx}{x {(lnx)}^{2}} = \int_{\ln 2}^{\infty} \frac{e^{t} dt}{e^{t} . t^{2}} = \int_{\ln 2}^{\infty} \frac{dt}{t^{2}} = {[- \frac{1}{t}]}_{\ln 2}^{\infty} = \frac{1}{\ln 2} < + \infty \Rightarrow this serie

is convergent \dots!