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As discussed in detail in that OP Find the limit of $\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { \sqrt { { n }^{ 2 } - { k }^{ 2 } } } { { n }^{ 2 } } } } $, using Riemann sum we obtain that $$\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { \sqrt { { n }^{ 2 } - { k }^{ 2 } } } { { n }^{ 2 } } } }=\lim _{ n\rightarrow \infty }\sum _{ k=1 }^{ n }\frac{1}{n}\sqrt{1-\frac{k^2}{n^2}}=\int_0^1\sqrt{1-x^2}{dx}=\frac{\pi}4$$

and using binomial expansion it can be shown that (for details refer to that answer)

$$\sum _{ k=1 }^{ n }\frac{1}{n}\sqrt{1-\frac{k^2}{n^2}}\to\sum_{j=0}^\infty \frac{(-1)^j\binom{\frac12}{j}}{{2j+1}}$$

and therefore the following holds

$$\sum_{j=0}^\infty \frac{(-1)^j\binom{\frac12}{j}}{{2j+1}}=\frac{\pi}4$$

Is there a way to show that result independently from the derivation by Riemann sum and the related integral?

Note that by the following identity the sum can also be written as

$$\sum_{j=0}^\infty \frac{\binom{2j}{j}}{2^{2j}(1-4j^2)}=\frac{\pi}4$$

and numerical extimations seem to confirm that sums hold (see sum 1 and sum 2).

Thanks in advance for your attention and for any help or suggestion!

user
  • 154,566

3 Answers3

4

A hypergeometric sum is what it is, in your case $$ S=\frac{1}{2}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\left[-\sqrt{1-x}+\frac{\arcsin\sqrt{x}}{\sqrt{x}}\right]_{x=1}=\frac{\pi}{4}$$ by standard Maclaurin series. How to equivalently represent the LHS stands up to you. Clearly

$$ S = \int_{0}^{1}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]\frac{x^{2n}}{1-2n}\,dx=\int_{0}^{1}\sqrt{1-x^2}\,dx$$ is related to the area of the unit circle via elementary integrals, not meaning that the only way for proving the above identity is to employ elementary integrals or Riemann sums. On the other hand the manipulation above is probably the most straightforward method to derive $S=\frac{\pi}{4}$. If you like a convoluted one, you may use $\frac{1}{4^n}\binom{2n}{n}=(-1)^n P_{2n}(0)$ together with $$ \frac{1}{\sqrt{(1+u)^2-4ux}}=\sum_{n\geq 0}u^n P_n(2x-1)$$ to derive just the same. Yet another way is to apply the binomial transform to $$ \frac{\pi}{4}=\arctan(1)=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}.$$

Jack D'Aurizio
  • 353,855
  • Not very straightforward answer then, at least for me! Very stimulating to go deeper into these topics. So many ways to get the result. Thanks a lot! – user Sep 01 '18 at 14:16
4

Using a CAS and simplifying the result to show the limit

$$I_p=\sum_{j=0}^p \frac{(-1)^j\binom{\frac12}{j}}{{2j+1}}=\frac \pi 4+(-1)^p\frac{ \binom{\frac{1}{2}}{p+1}}{2 p+3}\, _3F_2\left(1,p+\frac{1}{2},p+\frac{3}{2};p+2,p+\frac{5}{2};1\right)$$ The second term decreases very slowly to $0$ when $p$ increases as shown below $$\left( \begin{array}{cc} p & \text{second term} \\ 10 & 0.00274285 \\ 20 & 0.00100926 \\ 30 & 0.00055684 \\ 40 & 0.00036415 \\ 50 & 0.00026163 \\ 60 & 0.00019957 \\ 70 & 0.00015868 \\ 80 & 0.00013007 \\ 90 & 0.00010913 \\ 100 & 0.00009326 \\ 200 & 0.00003311 \\ 300 & 0.00001805 \\ 400 & 0.00001173 \\ 500 & 0.00000840 \\ 600 & 0.00000639 \\ 700 & 0.00000507 \\ 800 & 0.00000415 \\ 900 & 0.00000348 \\ 1000 & 0.00000297 \\ 2000 & 0.00000105 \\ 3000 & 0.00000057 \\ 4000 & 0.00000037 \\ 5000 & 0.00000027 \end{array} \right)$$

Claude Leibovici
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2

Use $$\sqrt{1+x}=\sum_{j=0}^\infty \binom{\frac12}{j}x^n$$ and let $x=-t^2$ then $$\sum_{j=0}^\infty \frac{(-1)^j\binom{\frac12}{j}}{{2j+1}}=\int_0^1\sqrt{1-t^2}\ dt=\frac{\pi}4$$

Nosrati
  • 29,995
  • There is a typo for $x^n \to x^j$. – user Sep 01 '18 at 12:57
  • Sorry but maybe I wasn't sufficiently clear in formulating the OP. I'm aware about the equivalence with integral and I'm not asking how to derive the equality by integral and series but I'm asking for an alternative derivation without Riemann sum and related integral. I add that point to the Op to make that more clear. Anyway thanks for your answer. – user Sep 01 '18 at 13:35