calculating a sequence sum with different values at odd and even

Question

Well that's supposed to be an easy one: given the sequence of

$$\left(\frac{3}{8}\right)^1+\left(\frac{3}{8}\right)^1\left(\frac{2}{7}\right)^1+\left(\frac{3}{8}\right)^2\left(\frac{2}{7}\right)^1+\left(\frac{3}{8}\right)^2\left(\frac{2}{7}\right)^2+\dots$$ How can I calculate this infinite sum? I know $\sum _{n=1}^{\infty }\:\left(\frac{3}{8}\right)^n=\frac{3}{5}$ and that $\sum _{n=1}^{\infty }\:\left(\frac{2}{7}\right)^n=\frac{2}{5}$.

Haven't played with sequences for a while so a bit help would be great. thanks

Answer 1

Since all the terms are positive you can rearrange and make groups (kind of commutative and associative properties).

Then your sum is $$\frac38\sum_{k=0}^\infty\left(\frac38\cdot\frac27\right)^n+\frac38\cdot\frac27\sum_{k=0}^\infty\left(\frac38\cdot\frac27\right)^n=\frac{27}{56}\cdot\frac1{1-\dfrac{6}{56}}=\frac{27}{50}$$

Remark: Under some conditions you can rearrange infinite sums even if some terms are negative. Search for absolutely convergent series.

Answer 2

You can split the sum into two parts: the elements where the exponents are the same and the elements in which they are not. This way you get

\begin{align} &\sum_{i=1}^\infty \left ( \frac{3\cdot 2}{8\cdot 7}\right )^i + \sum_{i=1}^\infty \left ( \frac{3}{8}\right )^i\left ( \frac{2}{7}\right )^{i-1} \\=&\sum_{i=1}^\infty \left ( \frac{3\cdot 2}{8\cdot 7}\right )^i + \frac{3}{8}\sum_{i=0}^\infty \left ( \frac{3\cdot 2}{8\cdot 7}\right )^i \\=&\sum_{i=1}^\infty \left ( \frac{3\cdot 2}{8\cdot 7}\right )^i + \frac{3}{8}\left (1+ \sum_{i=1}^\infty \left ( \frac{3\cdot 2}{8\cdot 7}\right )^i\right ) \end{align}

So the only sum you need to compute is $$ \sum_{i=1}^\infty \left ( \frac{3\cdot 2}{8\cdot 7}\right )^i. $$

You can use the geometric series formula to get $$ \sum_{i=1}^\infty \left ( \frac{3\cdot 2}{8\cdot 7}\right )^i = \frac{1}{1-3/28} -1= \frac{3}{25}, $$ and now you can easily obtain value of your sum.

Answer 3

Note the pattern for exponents: $1,1,2,2,3,3,... $

Hence: $$\left(\frac{3}{8}\right)^1+\left(\frac{3}{8}\right)^1\left(\frac{2}{7}\right)^1+\left(\frac{3}{8}\right)^2\left(\frac{2}{7}\right)^1+\left(\frac{3}{8}\right)^2\left(\frac{2}{7}\right)^2+\left(\frac{3}{8}\right)^3\left(\frac{2}{7}\right)^2+\cdots=\\ \color{red}{\frac38}+\frac{3\cdot 2}{8\cdot 7}+\color{red}{\frac38\cdot \frac{3\cdot 2}{8\cdot 7}}+\left(\frac{3\cdot 2}{8\cdot 7}\right)^2+\color{red}{\frac38\cdot \left(\frac{3\cdot 2}{8\cdot 7}\right)^2}+\left(\frac{3\cdot 2}{8\cdot 7}\right)^3+\cdots =\\ \frac38\left(\color{red}1+\color{red}q+\color{red}{q^2}+\cdots\right)+q+q^2+\cdots=...$$ where $q=\frac{3\cdot 2}{8\cdot 7}$. Can you finish?

The answer is:

$S=\frac{27}{50}$.