# Let a general term $T_n$ be defined as $T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$

Let a general term $T_n$ be defined as

$$T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$$

Then prove that $\lim_{n\to\infty}(T_1 + T_2 +\cdots+T_n) \lt \frac{4}{27}.$

I tried finding pattern between terms .. $T_1=\frac{1}{9} , \frac{T_2}{T_1}=(\frac{2}{5})^2, \frac{T_3}{T_2}=(\frac{3}{7})^2$ but could not think more of how to get a bound on the series. Any help is appreciated.

$$\dfrac{T_m}{T_{m-1}}=\left(\dfrac m{2m+1}\right)^2<\left(\dfrac m{2m}\right)^2=\dfrac14$$ for $m>0$

$$\sum_{r=1}^\infty T_r<\sum_{r=1}^\infty T_1\left(\dfrac14\right)^{r-1}=\dfrac{\dfrac19}{1-\dfrac14}=?$$

• how it is $\lt \frac{1}{4}$?? – Abhash Jha Apr 19 '17 at 8:49
• @AbhashJha, Please find the updated answer – lab bhattacharjee Apr 19 '17 at 8:50
• +1. Splendid. I arrived to a closed form for the sum $\left(~\mbox{namely,}\ 4\int_{0}^{\pi/6}{t \over \,\sqrt{1 - 4\sin^{2}\left(\,t\,\right)}\,}\,\mathrm{d}t - 1 ~\right)$. I couldn't go further. As Einstein said: "Keep it simple...". – Felix Marin Apr 21 '17 at 0:59

I do not think $\sum_{n\geq 1}T_n$ has a closed form, but such inequality can be improved a bit.

We have $$T_n = \left(\frac{n!}{(2n+1)!!}\right)^2 = \left(\frac{n!(2n)!!}{(2n+1)!}\right)^2 = \left(\frac{n!^2 2^n}{(2n+1)!}\right)^2 = \frac{4^n}{(2n+1)^2 \binom{2n}{n}^2}$$ and we may borrow a couple of useful lemmas from this answer: $$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx \tag{1}$$ $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac{1}{2}\sum_{n\geq 1}\frac{4^n x^{2n-1}}{n\binom{2n}{n}},\qquad \arcsin^2(x)=\frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}{n^2\binom{2n}{n}}\tag{2}$$ to derive: $$\sum_{n\geq 1}\frac{x^{2n+1}}{(2n+1)\binom{2n}{n}}=-x+\frac{4}{\sqrt{4-x^2}}\arcsin\frac{x}{2}\tag{3}$$ By $(1)$ and $(3)$ we get: $$\sum_{n\geq 1}T_n = \frac{1}{9}\cdot\phantom{}_3 F_2\left(1,2,2;\frac{5}{2},\frac{5}{2};\frac{1}{4}\right)=\int_{0}^{\pi/2}\left[-\sin(x)+\frac{4}{\sqrt{4-\sin^2 x}}\arcsin\frac{\sin x}{2}\right]\,dx$$ from which: $$\sum_{n\geq 1}T_n=-1+\int_{0}^{1}\frac{4\arcsin\frac{x}{2}}{\sqrt{(4-x^2)(1-x^2)}}\,dx=-1+\int_{0}^{\pi/6}\frac{4x}{\sqrt{1-4\sin^2 x}}\,dx \tag{4}$$ and by convexity the LHS of $(4)$ is bounded by $$-1+\int_{0}^{1}\frac{4\cdot\left[\frac{x}{2}+\left(\frac{\pi}{6}-\frac{1}{2}\right)x^3\right]}{\sqrt{(4-x^2)(1-x^2)}}\,dx = \frac{6-4\pi -9\log(3)+5\pi \log(3)}{6}<\color{red}{\frac{17}{127}}.\tag{5}$$ $(4)$ also provides a decent lower bound: $$\sum_{n\geq 1}T_n \geq -1+\int_{0}^{1}\frac{2x}{\sqrt{(1-x^2)(4-x^2)}}\,dx = \color{red}{-1+\log(3).}\tag{6}$$

• wait its not (2n+1)! as you have written in first step – Abhash Jha Apr 19 '17 at 9:07
• @AbhashJha: eh? $(2n+1)!! = (2n+1)\cdot(2n-1)\cdots 1$. – Jack D'Aurizio Apr 19 '17 at 9:08
• please see the denominator of $T_n$ in asked question – Abhash Jha Apr 19 '17 at 9:33
• @AbhashJha: I see it, it is the square of $1\cdot 3\cdots (2n+1)$, i.e. the square of $(2n+1)!!$. – Jack D'Aurizio Apr 19 '17 at 9:34
• @AbhashJha: are you aware of the notation for the double factorial? en.wikipedia.org/wiki/Double_factorial – Jack D'Aurizio Apr 19 '17 at 9:35

An additional method, though not as good as the above, use Stirling's Approximation for the relevant factorials. Then reduce. Should allow application of the ratio test.

https://en.wikipedia.org/wiki/Stirling%27s_approximation