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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.

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    $\begingroup$ You assert that in both cases "the area under the graph is finite." This is not true in the case of $f(x)$. A more familiar example is $\int_1^\infty \frac{1}{x}\,dx$. The function goes to $0$, but the area under the graph is not finite. $\endgroup$ Apr 8 '16 at 20:17
  • $\begingroup$ Visually though, why isn't f(x) finite? We already found using the divergent test that the limit of the function is 0 as x reaches infinity, so shouldn't the area also be finite? $\endgroup$ Apr 8 '16 at 20:32
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    $\begingroup$ We can compute $\int_2^M \frac{1}{x\sqrt{\ln x}}\,dx$. It turns out to be $2\sqrt{\ln M}-2\sqrt{\ln 2}$. This $\to\infty$ as $M\to\infty$. $\endgroup$ Apr 8 '16 at 20:40
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    $\begingroup$ What you're calling the "divergent test" is usually called (in elementary calculus, in the U.S.) the $n$th term divergence test. As to why the area should also be finite, it all comes down to adding more and more things that get smaller and smaller. If the "more and more" grows quickly enough compared to how small the things are getting, or equivalently if the "smaller and smaller" proceeds at a slow enough rate as you get more and more of them, then the resulting sum will not be finite. In short, a large number of very small numbers can still have a large sum. $\endgroup$ Apr 8 '16 at 20:45
  • $\begingroup$ Example :$1/2+1/3+1/4+1/5+1/6+1/7+1/8+...>$ $1/2 +1/4+1/4+1/8+1/8+1/8+1/8+...$ $=1/2+1/2+1/2+...=\infty.$ So if $\int_n^{n+1}f(x)\;dx \geq 1/n$ for $n\geq 2$ then $\int_2^{\infty}f(x)\;dx =\infty.$ $\endgroup$ Apr 8 '16 at 22:32
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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.

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