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.

  • 2
    $\begingroup$ Do you know geometric series such as 1+1/2+1/4...=2 $\endgroup$ – aman Mar 25 '19 at 12:24
  • $\begingroup$ And do you know the formula in general? $\endgroup$ – aman Mar 25 '19 at 12:24
  • $\begingroup$ If not, check on wikipedia or study it elsewher or you won't understand this sum. $\endgroup$ – aman Mar 25 '19 at 12:29
  • $\begingroup$ 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. $\endgroup$ – user647486 Mar 25 '19 at 12:32

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.

  • $\begingroup$ thanks! your answer is so helpful! and what does 's.t.' mean in 'For any 𝑥∈ℝ s.t. |𝑥|<1 we have' $\endgroup$ – shi95 Mar 25 '19 at 14:06
  • $\begingroup$ Oh sorry. s.t. is a shorthand notation for such that... our university professors use this notation. If you want to accept my answer click on the tick mark on the left side where two arrows appears, or you can vote my question up or can do both... in each case I will get reputation which will help me a lot. Thanks! $\endgroup$ – Sujit Bhattacharyya Mar 25 '19 at 15:27

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