I am having a bit of confusion doing the divergence and integral tests, specifically when I am trying to visualize the functions to get a better idea of why the methods work. For example, take the two functions \begin{align} f(x) &= \frac{1}{x\sqrt{\,\ln x}}\\[6px] g(x) &= \frac{2^{1/x}}{x^2} \end{align} Divergence test fails in both, as both sequences converge to $0$ as $x$ reaches infinity. However, when we use the convergence test where the bounds are $[2, \infty)$ for $f(x)$ and $[1, \infty)$ for $g(x)$, we find that:
Improper integral for $f(x)$ reaches infinity, therefore the series is divergent.
Improper integral for $g(x)$ reaches a value, therefore the series is convergent.
My question is, since both sequences converge to $0$, shouldn't the improper integral also converge to a number for both $f(x)$ and $g(x)$ since the "area under the graph" is finite (sequence converges to $0$)? Not sure where I went wrong here, any help is appreciated.
The integral of $f$ over $[2,\infty)$ can be explicitly computed; with the substitution $t=\ln x$, we have $$ \int\frac{1}{x\sqrt{\,\ln x}}\,dx=\int t^{-1/2}\,dt=\frac{\sqrt{t}}{2}+c =\frac{\sqrt{\,\ln x}}{2}+c $$ so $$ \lim_{t\to\infty}\int_2^t \frac{1}{x\sqrt{\,\ln x}}\,dx= \lim_{t\to\infty}\left(\frac{\sqrt{\,\ln t}}{2}-\frac{\sqrt{\,\ln 2}}{2}\right)=\infty $$ Thus the integral of $f$ over $[2,\infty)$ does not converge.
The second integral can instead be bounded above: if $x>1$, then $1/x<1$ and so $2^{1/x}<2$; therefore $$ 0<\frac{2^{1/x}}{x^2}<\frac{2}{x^2} $$ and $$ \lim_{t\to\infty}\int_{1}^{t}\frac{2}{x^2}\,dt =\lim_{t\to\infty}\left(-\frac{2}{t}+\frac{2}{1}\right)=2 $$ Therefore the integral of $g$ over $[2,\infty)$ is convergent.