Question #175892

does the series ∑ 1/n⋅[1+(ln n)2] converge or diverge

n=1 below ∑

∞ on top of ∑






1
Expert's answer
2021-03-30T08:04:39-0400

n=11n(1+(ln(n))2)\sum _{n=1}^{\infty \:}\frac{1}{n}\left(1+\left(\ln \left(n\right)\right)^2\right)

As an=1n(1+(ln(n))2);ax=1x(1+(ln(x))2)As \space a_n=\frac{1}{n}\left(1+\left(\ln \left(n\right)\right)^2\right) ; a_x=\frac{1}{x}\left(1+\left(\ln \left(x\right)\right)^2\right)

L=11x(1+(ln(x))2)=lnx+13ln3(x)+C1=L= \int_1^\infin \frac{1}{x}\left(1+\left(\ln \left(x\right)\right)^2\right)=\ln \left|x\right|+\frac{1}{3}\ln ^3\left(x\right)+C|_1^\infin =\infin

Hence n=11n(1+(ln(n))2)\sum _{n=1}^{\infty \:}\frac{1}{n}\left(1+\left(\ln \left(n\right)\right)^2\right) diverges


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