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.
$\endgroup$ 44 Answers
$\begingroup$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 $\pm1$).
$\endgroup$ $\begingroup$- 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 $S_2$.
$\endgroup$ 2 $\begingroup$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
$\endgroup$ $\begingroup$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
$$a_0 \cdot \dfrac{1}{1-q} = 11 \cdot \dfrac{1}{1-\dfrac{2}{11}} = \dfrac{121}{9}$$
