Question

If $a={\sum }_{n=1}^{\infty }\left(\frac{2n}{(2n-1)!}\right)$ and $b={\sum }_{n=1}^{\infty }\left(\frac{2n}{(2n+1)!}\right)$. Find value of [a] + [3b] where [ ] denotes greatest integer function or integral value of x.

Answer 1

$a={\sum }_{n=1}^{\infty }\frac{2n}{(2n+1)!}=e=2.71828b={\sum }_{n=1}^{\infty }\frac{2n}{(2n+1)!}=\frac{1}{e}=0.36787\left[a\right]+\left[3b\right]=\left[2.71828\right]+0[3\times 0.3678]=2+1=3\therefore \left[a\right]+\left[3b\right]=3$