So, a friend of mine told me the professor in one of his classes had computed the sum $\displaystyle \sum\limits_{n = 1}^{+\infty} n \cdot 0.3^{n - 1}$ by replacing (?) the summation with an integral. However, my friend had not written it down and could not recall anything of the procedure.
If I were to solve it, I'd do something like: $$ \sum\limits_{n = 1}^{+\infty} n \cdot 0.3^{n - 1} = [(1 + 0.3 + 0.3^2 + \ldots) + (0.3 + 0.3^2 + 0.3^3 + \ldots) + (0.3^2 + 0.3^3 + 0.3^4 + \ldots)] = \sum\limits_{n = 1}^{+\infty}\sum\limits_{i = n}^{+\infty} 0.3^{i} = \sum\limits_{n = 1}^{+\infty} \frac{0.3^{n - 1}}{1 - 0.3} = \frac{1}{0.7} \cdot \sum\limits_{n = 1}^{+\infty} 0.3^{n - 1} = \frac{1}{0.7} \cdot \frac{1}{1 - 0.3} = \frac{1}{0.49} \approx 2.04 $$
which I think is a correct result. So, how would this be solved with the help of integration?
Hint: for $\;|x|<1\;$ we have
$$f(x):=\frac1{1-x}=\sum_{n=0}^\infty x^n\implies f'(x)=\sum_{n=1}^\infty nx^{n-1}$$