0
$\begingroup$

I know that in order to apply the integral test for convergence or divergence a function $f(x)$ must be positive, continuous, and decreasing. However, I was wondering if $$ \int_{1}^{\infty}f(x)\, dx\, =\, \sum_{n=1}^{\infty}f(x) $$

because finding the integral of the function would essentially be finding the sum of the function in series itself?

Thank you!

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ No i don't believe it would, an easy way to test it would be to choose a decreasing convergent geometric series, apply the integral test, and then compare the answer from the integral test to the result of the geometric series (which is $\frac{a}{1-r}$ in case you forgot). I'm sure a more knowledgeable user can give you a proper proof. $\endgroup$ – helpmeh Apr 2 '16 at 18:58
  • $\begingroup$ @helpmeh Oh wow, I never thought of doing that! Lemme try $\endgroup$ – michaelchang64 Apr 2 '16 at 18:59
1
$\begingroup$

No, they are not equal. Consider the function $f(x)=\dfrac{1}{x^2}$. We have $$\int_1^\infty\frac{1}{x^2}\, dx=1$$ but $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.