Series Integration

To determine that $S_n =\sum_{n=1}^{\infty} nx^{n-1}$ converges, the $ratio$ $test$ can be implemented: $\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}| = \lim_{n\to\infty} |\frac{(n+1)x^{n+1-1}}{nx^{n-1}}| = \lim_{n\to\infty} |\frac{n+1}{n} \frac{x^n}{x^{n-1}}| \approx \lim_{n\to\infty} |x^{n-n+1}| = |x| < 1$ Thus, $S_n$ converges as long as $|x|<1$ , but the question now is to what does $S_n$ converge to?

The following is a well known $geometric$ $series$ that converges: $\sum_{n=1}^{\infty} x^n = \frac{1}{1-x}$ Notice that $\frac{d}{dx} x^n = nx^{n-1}$ ; therefore, $S_n =\sum_{n=1}^{\infty} nx^{n-1} = \frac{d}{dx}\sum_{n=1}^{\infty} x^n = \frac{d}{dx}\frac{1}{1-x}=\boxed{\frac{1}{(1-x)^2}}$

Determining $D(x)$ using $u$ - $substitution:$ $D(x)=\int S_ndx = \int \frac{1}{(1-x)^2} = -\int\frac{du}{u^2} = \frac{-1}{-1u} +c = \boxed{\frac{1}{1-x}+c}$

Finally, evaluating $J(x)$ at $x=0:$ $J(0) = D(0) - c = \frac{1}{1-0} +c-c = \frac{1}{1} = \boxed{1}$

$Note:$ We can abuse the relationship of integrals and derivatives to quickly determine $J(x):$ $J(x)=\int(\sum_{n=1}^{\infty} nx^{n-1})dx -c = \int(\frac{d}{dx}\sum_{n=1}^{\infty} x^n)dx-c =\int (\frac{d}{dx}\frac{1}{1-x})dx-c \\ J(x)=\frac{1}{1-x}+c-c= \boxed{\frac{1}{1-x}}$