Finding the infinite series: $3 \cdot \frac{9}{11} + 4 \cdot \left(\frac{9}{11}\right)^2 + 5 \cdot \left(\frac{9}{11}\right)^3 + \cdots$? I'm trying to find the infinite sum that is defined by:
$$
3 \cdot \frac{9}{11} + 4 \cdot \left(\frac{9}{11}\right)^2 + 5 \cdot \left(\frac{9}{11}\right)^3 + \cdots
$$
However, I do not know of any known formula to do this. Am I missing something really simple? Thanks!
 A: For $|r|<1$ we have $$\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}$$
Then, by taking derivatives respect to $r$
$$\sum_{k=0}^{\infty}kr^{k-1}=\frac{1}{(1-r)^2}\quad\implies\quad \frac{1}{r}+2+\sum_{k=3}^{\infty}kr^{k-2}=\frac{1}{r(1-r)^2}$$
Then
$$\sum_{k=3}^{\infty}kr^{k-2}=\frac{1}{r(1-r)^2}-\frac{1}{r}-2$$
Thus
$$\sum_{k=3}^{\infty}k\left(\frac{9}{11}\right)^{k-2}=\frac{1}{\frac{9}{11}\left(\frac{2}{11}\right)^2}-\frac{1}{\frac{9}{11}}-2=\frac{1331-44-72}{36}=\boxed{\color{blue}{\frac{135}{4}}}$$
A: The ideas in Mario G’s answer are very good ones to learn, but there are also ways to tackle the problem without calculus. You can easily test that the series is absolutely convergent, so we can rearrange terms pretty much at will. Now
$$\begin{align*}
3\left(\frac9{11}\right)+4\left(\frac9{11}\right)^2+5\left(\frac9{11}\right)^3+\ldots&=\sum_{n\ge 1}(n+2)\left(\frac9{11}\right)^n\\
&=\sum_{n\ge 1}n\left(\frac9{11}\right)^n+2\sum_{n\ge 1}\left(\frac9{11}\right)^n\\
&=\sum_{n\ge 1}\sum_{k=1}^n\left(\frac9{11}\right)^n+2\sum_{n\ge 1}\left(\frac9{11}\right)^n\\
&=\sum_{k\ge 1}\sum_{n\ge k}\left(\frac9{11}\right)^n+2\sum_{n\ge 1}\left(\frac9{11}\right)^n\;,
\end{align*}\tag{1}$$
and for each $k$ the summation $\sum_{n\ge k}\left(\frac9{11}\right)^n$ is just a convergent geometric series. In particular,
$$\sum_{n\ge k}\left(\frac9{11}\right)^n=\frac{\left(\frac9{11}\right)^k}{1-\frac9{11}}=\frac{11}2\left(\frac9{11}\right)^k\;,$$
and the desired sum is
$$\sum_{k\ge 1}\frac{11}2\left(\frac9{11}\right)^k+2\sum_{n\ge 1}\left(\frac9{11}\right)^n=\left(\frac{11}2+2\right)\sum_{k\ge 1}\left(\frac9{11}\right)^k=\frac{15}2\cdot\frac{11}2\cdot\frac9{11}=\frac{135}4\;.$$
If you’re not used to working with summations, the calculation in $(1)$ may look rather mysterious. If you write it out longhand in the right way, though, it becomes quite straightforward. For convenience I’ll replace $\frac9{11}$ by $r$. Then you can break up the sum in two dimensions:
$$\begin{array}{ccc}
3r&+&4r^2&+&5r^3&+&6r^4&+&\ldots\\ \hline
r&+&r^2&+&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 1}r^n&=&\frac{\color{red}r}{1-r}\\
+&&+&&+&&+\\
r&+&r^2&+&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 1}r^n&=&\frac{\color{red}r}{1-r}\\
+&&+&&+&&+\\
r&+&r^2&+&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 1}r^n&=&\frac{\color{blue}r}{1-r}\\
&&+&&+&&+\\
&&r^2&+&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 2}r^n&=&\frac{\color{blue}{r^2}}{1-r}\\
&&&&+&&+\\
&&&&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 3}r^n&=&\frac{\color{blue}{r^3}}{1-r}\\
&&&&&&+\\
&&&&&&r^4&+&\ldots&=&\sum_{n\ge 4}r^n&=&\frac{\color{blue}{r^4}}{1-r}\\ \hline
&&&&&&&&&&&&\frac1{1-r}\left(\color{red}{2r}+\color{blue}{\sum_{n\ge 1}r^n}\right)\\
&&&&&&&&&&&&=\frac1{1-r}\left(2r+\frac{r}{1-r}\right)
\end{array}$$
The original series is organized by summing the columns of this array first and then adding up the column sums; when I reversed the order of summation in the first term at the last step of $(1)$, I was switching to summing the rows first and then adding up the row sums.
A: You can write the series as 
$$\sum_{n=3}^{\infty}n\left(\frac{9}{11}\right)^{n-2}$$
By ratio test you would find that this series converge and you can find the sum by using geometric series.
