Can We Treat It As An Integral?

Another possible solution is if we let $f(x)=\frac { 1 }{ (\ln { (x) } )^{ m } }$ the sum becomes $\sum _{ x=2 }^{ \infty }{ f(x) }$ . Because $f(x)$ is positive, continuous, and decreasing on $[2,\infty )$ for positive integers $m$ , we can say that an approximation of the area under the curve $f(x)$ on $[2,\infty)$ which looks at sub-intervals of length $1$ and approximates the average value of $f(x)$ on that sub-interval by the left-most value of $f$ on it must be greater than the actual area under the curve. That is, $(1)f(2)+(1)f(3)+(1)f(4)+...=\sum _{ x=2 }^{ \infty }{ f(x) } >\int _{ 2 }^{ \infty }{ f(x)dx }$ . If we let $u=\ln { (x) }$ so that the lower bound becomes $u=\ln { (2) }$ and the upper bound remains ${ \infty }$ , and so that $\frac { du }{ dx } =\frac { 1 }{ x } \rightarrow xdu=dx\rightarrow e^{ u }du=dx$ , this shows that $\int _{ 2 }^{ \infty }{ f(x)dx } =\int _{ 2 }^{ \infty }{ \frac { 1 }{ (\ln { (x) } )^{ m } } dx } =\int _{ \ln { (2) } }^{ \infty }{ \frac { e^{ u } }{ u^{ m } } du }$ . Regardless of the value of $m$ , $\lim _{ u\rightarrow \infty }{ \frac { e^{ u } }{ u^{ m } } } =\infty$ , so that this integral and the original sum itself must diverge regardless of the value of $m$