How to justify the convergence of series $\left(\frac{1}{2}\right)^{2(2k-1)}$?

Question

I see this formula somewhere in a book, though the book doesn't provide the justification.

$$ \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{2(2k-1)} = \frac{4}{15} $$

Any clue would be appreciated.

Answer 1

Here is a way you may understand:

For any $x\in \Bbb R$ s.t. $|x|<1$ we have $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$ Proof is here.

Now in your case, let $\displaystyle S=\sum_{k=1}^\infty \frac{1}{4^{2k-1}}=\frac{1}{4}+\frac{1}{4^3}+\frac{1}{4^5}+\dots=\frac{1}{4}(1+\frac{1}{4^2}+\frac{1}{4^4}+\dots)=\frac{1}{4}\sum_{k=0}^\infty (\frac{1}{4^2})^k$

Now by above formula, [by putting $x=\frac{1}{4^2}$] $$S=\frac{1}{4}\frac{1}{1-\frac{1}{4^2}}=\frac{4}{15}$$

Hope this works.