If I have the following expression:
$\displaystyle\int_{0}^{\infty} f(x) dx = \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^3}$
Can I then deduce that $\int_{0}^{\infty} f(x) dx$ converges, because right hand side does?
If I have the following expression:
$\displaystyle\int_{0}^{\infty} f(x) dx = \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^3}$
Can I then deduce that $\int_{0}^{\infty} f(x) dx$ converges, because right hand side does?
Well...yes, of course! What you wrote says the integral equals a number , so yes: this means the integral converges.
If I were to say, for example, for a finite $C$
$\int^{\infty}_{0}f(x)\, dx=C < \infty \implies \int^{\infty}_{0}f(x)\, dx < \infty$
Thus
$\displaystyle\int_{0}^{\infty} f(x) \,dx=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^3}=\zeta(3)$
So the integral converges! You even know what the integral converges to.