<s>[INST] How to justify the convergence of series $\left(\frac{1}{2}\right)^{2(2k-1)}$? 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.
 [/INST] 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.
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