Originally Posted by

**arbolis** Today I spent some time about an infinite series. I came up with $\displaystyle \sum_{i=1}^{+\infty}\frac{1}{3^i}$. I know it converges (thanks to the integral test of convergence. $\displaystyle \int_1^{\infty}\frac{1}{3^x}dx=\frac{1}{3ln(3)}$). Now I don't understand why the value of the integral is not the value of the series. A decreasing function $\displaystyle f:[a,b]\rightarrow R$ has a definite integral $\displaystyle \int_a^b f(x) dx$ equal to $\displaystyle \frac{b-a}{n}\sum_{i=1}^n (t_i-t_{i-1})f(t_i)$. In our case $\displaystyle f(x)=\frac{1}{3^x}$ is decreasing. Now if $\displaystyle a=1$, $\displaystyle b$ and $\displaystyle n$ tends to $\displaystyle +\infty$ the integral becomes the infinite series I came up with, which is worth $\displaystyle \frac{1}{2}$ and not $\displaystyle \frac{1}{3ln(3)}$. Where is my error? Also, __if the integral I have and the infinite series I have are not equal, why can we use the integral test for convergence, while working on an infinite series?__ It would mean that if the series converges, then the integral does so, but their values are not equal... Strange. And of course, if an integral converges, so does its infinite series and their values are equal. But in my example I made an error, that's why it doesn't work.

Last question : is it possible to determine to what value converges the series **using integrals**? Thanks in advance.