Question

${lim}_{n\to \infty }{\sum }_{r=1}^n\frac{r}{1\times 3\times 5\times 7\times ...\times (2r+1)}$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. none of these

Answer 1

The correct option is A $\frac{1}{3}$

$T\left(r\right)=\frac{r}{1\times 3\times 5\times \cdots \times (2r+1)}$

$=\frac{2r+1-1}{2(1\times 3\times 5\dots (2r+1\left)\right)}$

$=\frac{1}{2}(\frac{1}{1\times 3\times 5\dots (2r-1)}-\frac{1}{1\times 3\times 5\dots (2r+1)})$

$=-\frac{1}{2}[V\left(r\right)-V(r-1)]$

$\Rightarrow \sum _{r=1}^{n}T\left(r\right)=-\frac{1}{2}\left(V\right(n)-V(0\left)\right)$
as all other terms get cancelled out due to repetitive addition

$=\frac{1}{2}(1-\frac{1}{1\times 3\times 5\times \cdots \times (2n+1)})$

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{r}{1\times 3\times 5\times \cdots \times (2r+1)}$

$=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{2}(1-\frac{1}{1\times 3\times 5\times \cdots \times (2r+1)})$
$={lim}_{n\to \infty }\frac{1}{2}(1-\frac{1}{1\times 3\times 5\times \cdots \times (2n+1)})$

$=\frac{1}{2}$