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I was playing around with numbers and noticed that the following series is quite close to $\sqrt 2$... but not quite. So I have conjectured that $\sqrt 2$ is arguably the closest highest bound;

$$1+\frac 12\left(1-\frac{1}{2^2}\right)+\frac{1\cdot 3}{2\cdot 4^2}\left(\frac{1}{2}-\frac{1}{3^2}\right)+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6^3}\left(\frac{1}{3}-\frac{1}{4^2}\right)+\cdots < \sqrt{2}$$

Can this be proven? Is there a closed form of the infinite series? I... have no idea how to tackle this problem. I've never seen anything like this and have just made it up because I was bored in physics class, but after some iterations, methinks this series is convergent.

Are there any series that look like this, particularly for $\sqrt 2$, that may serve a good use of application to measuring the truth of this conjecture?

Thanks. :)

Mr Pie
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    If this series actually converges to $\sqrt{2}$, you could be a second Ramanujan. – Peter Feb 13 '20 at 23:52
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    @Peter I am flattered, but unlike Ramanujan, I don't believe in an elephant goddess :) – Mr Pie Feb 13 '20 at 23:55
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    @MrPie your infinite series is equal to $$1+\sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n-2)!!(2n)^n} \left( \frac{1}{n} - \frac{1}{(n+1)^2} \right)$$ where $n!!$ is the double factorial. I am pretty sure this series converges using the ratio test. – Mathphile Feb 14 '20 at 00:40
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    After running a PARI program your series seems to converge to $1.413925843729...$ which is less than $\sqrt{2}$ – Mathphile Feb 14 '20 at 04:13
  • @Mathphile given that $\sqrt 2 = 1.4142\ldots$ then the percentage difference is roughly $0.0203%$ which is very small. – Mr Pie Feb 14 '20 at 04:43
  • If you add $e^{-e\pi}$ to $1.4139258\ldots$ you get $1.4141213\ldots$ which is even closer to $\sqrt 2$. – Mr Pie Feb 14 '20 at 07:44

2 Answers2

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As @Mathphile pointed out in the comments, the series can be written as $$S=1+\sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n-2)!!\cdot(2n)^n} \left( \frac{1}{n} - \frac{1}{(n+1)^2} \right)=1+\sum_{n=1}^{\infty} \frac{a_nb_n}{2^{3n-2}}$$ where $$a_n=\frac{(2n-1)!}{n^n\cdot(n-1)!^2}\quad\text{and}\quad b_n=\frac{1}{n} - \frac{1}{(n+1)^2},$$ since $(2n-1)!!=(2n - 1)! / [2^{n - 1} \cdot (n - 1)!]$ and $(2n-2)!!=2^{n - 1} \cdot (n - 1)!$. Clearly the terms of $b_n$ form a decreasing sequence, and this also holds for $a_n$ since $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{2\left(\frac2n+\frac1{n^2}\right)}{1+\frac1n}\left(1-\frac1{n+1}\right)^n<1.$$ Thus $$S=1+\sum_{n=1}^{10}\frac{a_nb_n}{2^{3n-2}}+\sum_{m>10}\frac{a_{10}b_{10}}{2^{3m-2}}<1+0.414+10^{-14}<\sqrt2.$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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With a little complication (see below), we can prove the result by comparing your series to a well-known series $$\sqrt{2}=1+\sum_{n=1}^\infty\frac{(2n-1)!!}{4^nn!}$$

(see for example Infinite series for $ \sqrt 2 $ ). Let $n\geq 1$ and $$a_0=1,a_n= \frac{(2n-1)!!}{(2n-2)!!(2n)^n} \left( \frac{1}{n} - \frac{1}{(n+1)^2} \right) \\ b_0=1,b_n=\frac{(2n-1)!!}{4^nn!}$$ be terms of both series. We can prove that $a_n < b_n$ for $n > 1$. Writing up all the terms, simplifying and cancellation of double factorials, this boils down to showing $$ \frac{2}{n^{n-1}}\left(\frac{1}{n}-\frac{1}{(n+1)^2}\right) < 1. $$

The real difficulty is $a_1>b_1$, but we can work around this by comparing couple more terms together. Indeed, by direct evaluation (with aid of a computer) we see $$\sum_{1 \leq n \leq 10}a_n < \sum_{1 \leq n \leq 10}b_n.$$

Combining this with term-wise comparison described above, we finally get $$ \sum_{n\geq 0}a_n=a_0+\sum_{1 \leq n \leq 10}a_n+\sum_{n\geq 11}a_n < b_0+\sum_{1 \leq n \leq 10}b_n+\sum_{n\geq 11}b_n=\sum_{n\geq 0}b_n=\sqrt{2}. $$

Of course, it would be more elegant if we could somehow avoid the calculation of those $10$ terms, that I guess is something to think about.

Sil
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