does the series ∑ 1/n⋅[1+(ln n)2] converge or diverge
n=1 below ∑
∞ on top of ∑
∑n=1∞ 1n(1+(ln(n))2)\sum _{n=1}^{\infty \:}\frac{1}{n}\left(1+\left(\ln \left(n\right)\right)^2\right)∑n=1∞n1(1+(ln(n))2)
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)As an=n1(1+(ln(n))2);ax=x1(1+(ln(x))2)
L=∫1∞1x(1+(ln(x))2)=ln∣x∣+13ln3(x)+C∣1∞=∞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 =\infinL=∫1∞x1(1+(ln(x))2)=ln∣x∣+31ln3(x)+C∣1∞=∞
Hence ∑n=1∞ 1n(1+(ln(n))2)\sum _{n=1}^{\infty \:}\frac{1}{n}\left(1+\left(\ln \left(n\right)\right)^2\right)∑n=1∞n1(1+(ln(n))2) diverges
