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I am having some trouble with the infinite series $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln^2n}$ . I used the integral test and simplified it to $\int_{\ln 2}^b - \frac 1{\ln(n)}$ (implied infinity since improper integral). But this website:

http://www.math.northwestern.edu/~mlerma/courses/b17-99w/b17w99-1mid-ans.pdf

# 2 on the website has the same exact question that I have but instead of the answer on the website being bounded from $\ln(2)$ to $b$, its $2$ to $b$. Doesn't the boundaries have to change because of the bounds should be in terms of the $u$-substitution? So since $u = \ln(n)$. Plug in $\ln(2)$...

Teoc
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Nocturne
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2 Answers2

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You only change the limits of integration if you are trying to evaluate your integral in terms of your substitution. If you do a $u$-sub, integrate, then convert your antiderivative back before evaluating at the limits of integration, then you need the original limits.

The integral test says you need to evaluate $$\lim_{N \to \infty} \int_2^N \frac{dx}{x\ln^2(x)} = \lim_{N \to \infty}\left[ \frac{-1}{\ln(x)}\right]_2^N \\ = 0-\left(-\frac{1}{\ln(2)}\right) \\ = \frac{1}{\ln(2)}<\infty$$ so the sum is convergent by the integral test.

graydad
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  • I understand, but I have another question. Wouldn't that mean that there is no point in changing the limits of integration in terms of u since you would be converting the answer back to its original form to get the answer? I don't understand when I would try to use the integral boundaries in terms of u in the first place. – Nocturne Jan 28 '15 at 04:32
  • Sometimes the sub is very nasty, and it's just easier to convert the limits instead of undoing the sub after getting the antiderivative. It's just an option we can choose, but in either case the same answer will result. – graydad Jan 28 '15 at 04:33
  • @Nocturne If you are interested in an example, check out the sub I used in my answer here. http://math.stackexchange.com/questions/1100394/how-is-the-integral-int-frac1aemxbe-mx-dx-solved/1100437#1100437

    Had this been a definite integral, I would definitely have changed the limits instead of undoing the sub.

     – graydad Jan 28 '15 at 04:37
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Here's where you got a bit mixed up: $~\displaystyle\sum_{n=2}^{\infty}\frac1{n\ln^2n}~\approx~\int_2^\infty\frac{dx}{x\ln^2x}~=~\int_{\ln2}^\infty\frac{dt}{t^2}$

Lucian
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