# Would the sum after applying the integral test be equal to the sum of a series?

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!

• 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. – helpmeh Apr 2 '16 at 18:58
• @helpmeh Oh wow, I never thought of doing that! Lemme try – michaelchang64 Apr 2 '16 at 18:59

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 $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$