e Nature

I've got two paths to show:

• $1)$

We'll manipulate our popular expansion of e.

$1+\frac3{1!}+\frac5{2!}+\frac7{3!}+...$

$=1+\frac{1+2}{1!}+\frac{1+4}{2!}+\frac{1+6}{3!}+...$

$=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+...+2(\frac1{1!}+\frac2{2!}+\frac3{3!})+...)$

$=e+2(1+\frac1{1!}+\frac1{2!}+...)$

$=3e$

$Qt\pi$

• $2)$ (Alternative)

We'll use this relation: $\Sigma \frac1{(n-k)!} = e$

Check the n-th term :

$u_n=\frac{2n-1}{(n-1)!}$

$\quad \; =\frac{2(n-1)+1}{(n-1)!}$

$\quad \; =\frac2{(n-2)!}+frac1{(n-1)!}$

Hence,

$\Sigma u_n =2e+e$

$\quad \quad =3e$

$Qt\pi$