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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
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B
32
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C
12
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D
none of these
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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

$$ = \dfrac{1}{2} \Big( 1 - \dfrac{1}{1 \times 3 \times 5 \times \dots \times (2n + 1)} \Big)$$

$$\displaystyle  \lim_{n \rightarrow \infty} \sum_{r=1}^{n} \dfrac{r}{1 \times 3 \times 5 \times \dots \times (2r+1)}$$

$$\displaystyle  = \lim_{n \rightarrow \infty} \sum_{r=1}^n  \dfrac{1}{2} \Big(1 - \dfrac{1}{1 \times 3 \times 5 \times \dots \times (2r+1)} \Big)
 $$
$$ = \lim_{n \rightarrow \infty} \dfrac{1}{2} \Big( 1 - \dfrac{1}{1 \times 3 \times 5 \times \dots \times (2n + 1)} \Big)$$

$$= \dfrac{1}{2}$$

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