# Does sum_{n=2} 1 / (1 + n ( Ln(n) )^2) converges or diverges from n=2 to infinity?

Jun 17, 2016

${\sum}_{i = 2}^{\infty} \frac{1}{1 + n {\left({\log}_{e} n\right)}^{2}}$ is convergent

#### Explanation:

$1 + n {\left({\log}_{e} n\right)}^{2} > n {\left({\log}_{e} n\right)}^{2}$

also

$\frac{1}{1 + n {\left({\log}_{e} n\right)}^{2}} < \frac{1}{n {\left({\log}_{e} n\right)}^{2}}$

So if ${\sum}_{i = 2}^{\infty} \frac{1}{n {\left({\log}_{e} n\right)}^{2}}$ is convergent then

${\sum}_{i = 2}^{\infty} \frac{1}{1 + n {\left({\log}_{e} n\right)}^{2}}$ will be convergent

but ${\int}_{2}^{n} \frac{\mathrm{dx}}{x {\left({\log}_{e} x\right)}^{2}} \ge {\sum}_{i = 3}^{n + 1} \frac{1}{n {\left({\log}_{e} n\right)}^{2}}$

because $\frac{1}{x {\left({\log}_{e} x\right)}^{2}}$ is monotonically decreasing

and $\int \frac{\mathrm{dx}}{x {\left({\log}_{e} x\right)}^{2}} = - \frac{1}{\log} _ e \left(x\right)$

also

{ (lim_{x->oo}-1/log_e(x) = 0), (lim_{n->oo}1/( n (log_e n)^2)=0) :}

So

${\sum}_{i = 2}^{\infty} \frac{1}{1 + n {\left({\log}_{e} n\right)}^{2}}$ is convergent

Comparison between

${\int}_{2}^{n} \frac{\mathrm{dx}}{x {\left({\log}_{e} x\right)}^{2}}$ and ${\sum}_{i = 3}^{n + 1} \frac{1}{n {\left({\log}_{e} n\right)}^{2}}$