Interesting Sum!

We can prove by induction that $\displaystyle S_n = \sum_{k=1}^n \dfrac{k}{(2k+1)!!} = \dfrac{(2n+1)!!-1}{2(2n+1)!!}$ .

• For $n=1$ , $\displaystyle S_n = \dfrac{3!!-1}{2\times 3!!} = \dfrac{3-1}{2 \times 3} = \dfrac{1}{3}$ , which is true.
• Assume that the formula is true for all $n$ .
• Then for $n+1$ , we have:

$\begin{aligned} \quad \quad S_{n+1} & = S_n + \frac{n+1}{(2n+3)!!}\\ & = \frac{(2n+1)!!-1}{2(2n+1)!!} + \frac{n+1}{(2n+3)!!} \\ & = \frac{[(2n+1)!!-1](2n+3) + 2(n +1) }{2(2n+3)!!} \\ & = \frac{(2n+3)!!-2n-3 + 2n +2 }{2(2n+3)!!} \\ & = \frac{[2(n+1)+1]!!-1}{2[2(n+1)+1]!!} \text{, which is true.} \quad \blacksquare \end{aligned}$

Therefore,

$\begin{aligned} S_n & = \frac{(2n+1)!!-1}{2(2n+1)!!} \\ \Rightarrow S_\infty & = \lim_{n \to \infty} \frac{(2n+1)!!-1}{2(2n+1)!!} = \lim_{n \to \infty} \left(\frac{1}{2}- \frac{1}{2(2n+1)!!}\right) = \frac{1}{2} \\ \Rightarrow 2 S_\infty & = \boxed{1} \end{aligned}$

Consider a sequence $T_0 ,T_1 , T_2, ......... T_k$ given by

$T_k = \dfrac{1}{1 \cdot 3 \cdot 5 \cdot 7 \cdot...........(2k +1) }$

Notice that the given summation $S$ is $\displaystyle \sum_{k=1}^{ \infty}U_k$ where

$U_k = T_{k-1} - T_{k}$

Thus $\displaystyle S = \sum_{k=1}^{ \infty} {(T_{k-1} - T_{k})}$

$\implies S = ( T_0 - T_1 )+ ( T_1 - T_2 ) + (T_2 - T_3 ).............$

Also, $\displaystyle \lim_{n \rightarrow \infty }{ T_n } = 0$

$\implies \color{#3D99F6}{S = T_0 = 1 }$