Partial sums of $\frac{\pi}{2}=\sum_{n=0}^\infty \frac{(2n-1)!!}{2^n\cdot n!\cdot (2n+1) }$.

Question

Recently, I have found a formula for $\pi$. That is $$\frac{\pi}{2}=\sum_{n=0}^\infty \frac{(2n-1)!!}{2^n\cdot n!\cdot (2n+1) }$$

However, the problem arises when you take the partial sums. For instance, $$\sum_{n=0}^{50000} \frac{(2n-1)!!}{2^n\cdot n! \cdot (2n+1)} \approx 1.57294$$

$$\sum_{n=0}^{100000} \frac{(2n-1)!!}{2^n\cdot n! \cdot (2n+1)} \approx 1.57439$$

But $\dfrac{\pi}{2} \lt 1.5708$

In other words, it can't exceed $1.5708$. So, how can it have a sum of $1.57439$ after summing $10^5+1$ terms if the sign of the series doesn't alternate?

• All the calculations were performed by Wolfram|Alpha.

Answer 1

I think you made some mistake while calculating the sums on WolframAlpha

here I calculated the first sum and it is equal to approx $1.08652$

$$\sum_{n=0}^{50000} \frac{(2n-1)!!}{2^n\cdot n! \cdot (2n+1)!} \approx 1.08652$$

Here

$$\sum_{n=0}^\infty \frac{(2n-1)!!}{2^n\cdot n!\cdot (2n+1) }\approx 1.57294$$

Here the sum is exceeding because of floating point approximations and you can never guarantee that WolframAlpha will always give you correct answers.