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

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.

• Do you know geometric series such as 1+1/2+1/4...=2 – aman Mar 25 '19 at 12:24
• And do you know the formula in general? – aman Mar 25 '19 at 12:24
• If not, check on wikipedia or study it elsewher or you won't understand this sum. – aman Mar 25 '19 at 12:29
• You can multiply a partial sum by $1-\frac{1}{2^4}$, distribute the multiplication, and see how all terms cancel except the first and the last. The last term will be $\frac{1}{2^{2(2n+1)}}$, which tends to $0$. Therefore, the sum will be equal to the first term, divided by the factor that we added $1-\frac{1}{2^4}$. The choice of the factor is $1$ minus the ratio of consecutive terms. That is what makes the terms cancel. – user647486 Mar 25 '19 at 12:32

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 $$x=\frac{1}{4^2}$$] $$S=\frac{1}{4}\frac{1}{1-\frac{1}{4^2}}=\frac{4}{15}$$