can someone help me go through this question i cant figure it out, i need help asap
$\displaystyle \sum_{r=1}^{\infty} \frac{1}{3^{r-2} r} $
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{3^{n-2}n}\leq\sum_{n=1}^{\infty}\frac{1}{3^{n-2}}=\frac{9}{2}$
Now for the sum rewrite it as
$\displaystyle 9\sum_{n=1}^{\infty}\frac{\left(\frac{1}{3}\right) ^n}{n}$
Now it is known that
$\displaystyle \sum_{n=1}^{\infty}\frac{\left(x\in[-1,1)\right)^n}{n}=-\ln(1-x)$
So our sum is
$\displaystyle 9\ln\left(\frac{3}{2}\right)$
Sum of infinite geometric series:
$\displaystyle \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ for $\displaystyle |r| < 1$.
Integrate both sides (there are some technical issues regarding the justification of changing the order of operations):
$\displaystyle \sum_{n=0}^{\infty} \frac{r^{n+1}}{n+1} = - \ln (1 - r) + C$ for $\displaystyle |r| < 1$
Substitute $\displaystyle r = 0: ~ 0 = 0 + C \Rightarrow C = 0$
$\displaystyle \Rightarrow \sum_{n={\color{red}1}}^{\infty} \frac{r^n}{n} = - \ln (1 - r)$ for $\displaystyle |r| < 1$.