Infinite Sum

[man111]

If [tex]\displaystyle a_{n}=\bigg(\frac{n!}{1\cdot 3\cdot 5\cdots (2n+1)}\bigg)^2[/tex] .

Then proving the result[tex]\displaystyle \lim_{n\rightarrow\infty}\sum^{n}_{k=1}a_{k}<\frac{4}{27}[/tex]

[pal702004]

$\dfrac{a_n}{a_{n-1}}=\left(\dfrac{n}{2n+1}\right)^2<\dfrac 1 4$

$a_1=\dfrac 1 9 \Longrightarrow$

$\sum_{k=1}^{\infty} a_k<\sum_{k=1}^{\infty} \dfrac{1}{9\cdot 4^{k-1}}=\dfrac 1 9 \cdot \frac 4 3 =\dfrac{4}{27}$