The sum 1+(1+3)/(2!)+(1+3+5)/(3!)+(1+3+5+7)/(4!)+ upto terms, is equal to

Question

The sum $1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots$ upto $\infty$ terms, is equal to

Accepted Answer

The series given is:

$ 1 + \frac{1+3}{2!} + \frac{1+3+5}{3!} + \frac{1+3+5+7}{4!} + \ldots $

In general, the $ r $-th term of the series, denoted as $ T_r $, takes the form:

$ T_r = \frac{r^2}{r!} $

To simplify $ T_r $, we write it in terms of factorials:

$ T_r = \frac{r}{(r-1)!} = \frac{(r-1)+1}{(r-1)!} = \frac{1}{(r-2)!} + \frac{1}{(r-1)!} $

Thus, the series can be expressed as:

$ \sum\limits_{r=1}^{\infty} T_r = \sum\limits_{r=1}^{\infty} \left( \frac{1}{(r-2)!} + \frac{1}{(r-1)!} \right) $

This breaks into two parts:

$ = \sum\limits_{r=1}^{\infty} \frac{1}{(r-2)!} + \sum\limits_{r=1}^{\infty} \frac{1}{(r-1)!} $

Each part is a well-known series. Specifically:

$\sum\limits_{r=1}^{\infty} \frac{1}{(r-2)!} = e$, starting from the term where $ r-2 = 0$, which aligns with the expansion of $ e $.

$\sum\limits_{r=1}^{\infty} \frac{1}{(r-1)!} = e$, for the terms starting where $ r-1 = 0$.

Consequently, the sum of the entire series is:

$ e + e = 2e $

Solutions · 1