Finding the sum of the infinite series Hi I am trying to solve the sum of the series of this problem:
$$
11 + 2 + \frac 4 {11} + \frac 8 {121} + \cdots
$$
I know its a geometric series, but I cannot find the pattern around this. 
 A: *

*If your series  is


$$
S_1=11+11\times\sum_{n=1}^{\infty}\frac{2^n}{11^{n}}
$$
then you may use 

$$
x+x^2+\cdots+x^n+\cdots=\frac{x}{1-x}, \quad |x|<1, \tag1
$$ 

with $x=\dfrac2{11}$.


*

*If your series  is


$$
S_2=11+2+\sum_{n=1}^{\infty}\frac{4n}{11^{n}}
$$
then differentiating $(1)$ termwise and multiplying by $x$ you get

$$
x+2x^2+3x^3+\cdots+nx^n+\cdots=\frac{x}{(1-x)^2}, \quad |x|<1, \tag2
$$ 

giving, with $x=\dfrac1{11}$,
$$
\sum_{n=1}^{\infty}\frac{n}{11^{n}}=\frac{11}{100}
$$ 
It is now pretty easy to obtain $S_2$.
A: Q: What does it mean to say a series is geometric?A: It means there is a common ratio.${}\qquad{}$I.e. the number by which you multiply each term to get the next is the same in every case.
What do you multiply $11$ by to get $2$?  You multiply it by $\dfrac 2 {11}$.
What do you multiply that by to get $\dfrac 4 {11}$?  You multiply it by $\dfrac 2 {11}$.
What do you multiply that by to get $\dfrac 8 {121}$?  You multiply it by $\dfrac 2 {11}$.
The fact that the thing by which you multiply is always the same thing is what the word "common" means.
The common ratio is $r=\dfrac 2 {11}$.
Remember that the sum of a geometric series is $\dfrac a {1-r}$, where $a$ is the first term (provided the common ratio is between $\pm1$).
A: Hint
The geometric sequence can be rewritten as $\frac{1}{11^{-1}}, \frac{2}{11^0}, \frac{4}{11^1}, \frac{8}{11^2}...$
Notice the powers of two on top and powers of eleven on bottom
A: Your series is a geometric series with a common ration of $q =\dfrac{2}{11}$ and a first term $a_0$ equal to $11$.
$-1 < \dfrac{2}{11}< 1$ so the series is convergent.
Its sum is equal to
$$a_0 \cdot \dfrac{1}{1-q} = 11 \cdot \dfrac{1}{1-\dfrac{2}{11}} = \dfrac{121}{9}$$
