Question

# $$\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\dfrac{r}{1\times 3\times 5\times 7\times ...\times (2r+1)}$$ is equal to

A
13
B
32
C
12
D
none of these

Solution

## The correct option is A $$\dfrac{1}{3}$$$$T(r) = \dfrac{r}{1 \times 3 \times 5 \times \dots \times (2r + 1)}$$$$= \dfrac{2r+1-1}{2(1 \times 3 \times 5 \dots (2r + 1))}$$$$= \dfrac{1}{2} \Big( \dfrac{1}{1 \times 3 \times 5 \dots (2r-1)} - \dfrac{1}{1 \times 3 \times 5 \dots (2r+1)} \Big)$$$$= -\dfrac{1}{2} \Big[ V(r) - V(r-1) \Big]$$$$\displaystyle \Rightarrow \sum_{r=1}^n T(r) = -\dfrac{1}{2} (V(n) - V(0))$$as all other terms get cancelled out due to repetitive addition$$= \dfrac{1}{2} \Big( 1 - \dfrac{1}{1 \times 3 \times 5 \times \dots \times (2n + 1)} \Big)$$$$\displaystyle \lim_{n \rightarrow \infty} \sum_{r=1}^{n} \dfrac{r}{1 \times 3 \times 5 \times \dots \times (2r+1)}$$$$\displaystyle = \lim_{n \rightarrow \infty} \sum_{r=1}^n \dfrac{1}{2} \Big(1 - \dfrac{1}{1 \times 3 \times 5 \times \dots \times (2r+1)} \Big)$$$$= \lim_{n \rightarrow \infty} \dfrac{1}{2} \Big( 1 - \dfrac{1}{1 \times 3 \times 5 \times \dots \times (2n + 1)} \Big)$$$$= \dfrac{1}{2}$$Maths

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