Question

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Answer 1

Answer:

A) 1/2

Step-by-step explanation:

We need to evaluate the below infinite series

$\displaystyle \longrightarrow \sum \limits_{n = 1}^ \infty \dfrac{n}{3 \cdot5 \cdot7 ...(2 n + 1)}$

Consider the nth term of the given series,

$\displaystyle \implies T_n= \dfrac{n}{3 \cdot5 \cdot7 ...(2 n + 1)}$

${ \displaystyle \implies T_n= \frac{1}{2} \left[ \dfrac{2n}{3 \cdot5 \cdot7 ...(2 n + 1)} \right]}$

${ \displaystyle \implies T_n= \frac{1}{2} \left[ \dfrac{(2n + 1 ) - 1}{3 \cdot5 \cdot7 ...(2 n + 1)} \right]}$

${ \displaystyle \implies T_n= \frac{1}{2} \left[ \dfrac{(2n + 1 )}{3 \cdot5 \cdot7 ...(2 n + 1)} - \dfrac{1}{3 \cdot5 \cdot7 ...(2 n + 1)} \right]}$

${ \displaystyle \implies T_n= \frac{1}{2} \left[ \dfrac{(2n + 1 )}{3 \cdot5 \cdot7 ...(2n - 1) \cdot(2 n + 1)} - \dfrac{1}{3 \cdot5 \cdot7 ...(2 n + 1)} \right]}$

${ \displaystyle \implies T_n= \frac{1}{2} \left[ \dfrac{1}{3 \cdot5 \cdot7 ...(2n - 1)} - \dfrac{1}{3 \cdot5 \cdot7 ...(2 n + 1)} \right]}$

Now the sum of the series is given

${\implies S_n =\lim\limits_{n\to\infty} (T_1+T_2+T_3+...+T_n)}$

${\implies S_n =\lim\limits_{n\to\infty} \dfrac{1}{2} \left[\left\{\dfrac{1}1-\dfrac1{1.3}\right\}+\left\{\dfrac1{1.3}-\dfrac{1}{1.3.5}\right\}+...+\left\{\dfrac{1}{1.3.5...(2n-1)}-\dfrac{1}{1.3.5...(2n+1)}\right\}\right] }$

${\implies S_n =\lim\limits_{n\to\infty} \dfrac{1}{2} \left[\left\{\dfrac{1}1 \cancel{-\dfrac1{1.3}}\right\}+\left\{ \cancel\dfrac1{1.3}- \cancel\dfrac{1}{1.3.5}\right\} \cancel{+...+}\left\{ \cancel\dfrac{1}{1.3.5...(2n-1)}-\dfrac{1}{1.3.5...(2n+1)}\right\}\right] }$

${\implies S_n =\lim\limits_{n\to\infty} \dfrac{1}{2} \left[\left\{\dfrac{1}1 -\dfrac{1}{1.3.5...(2n+1)}\right\}\right] }$

${\implies S_n =\dfrac{1}{2} \left[\left\{\dfrac{1}1 -\dfrac{1}{ \infty}\right\}\right] }$

${\implies S_n =\dfrac{1}{2} \left[\left\{\dfrac{1}1 -0\right\}\right] }$

${\implies S_n =\dfrac{1}{2} \left[1\right] }$

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

Therefore, the required answer is,

$\underline{\boxed{\sum \limits_{n = 1}^ \infty \dfrac{n}{3 \cdot5 \cdot7 ...(2 n + 1)} = \frac{1}{2} }}$

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Answer 2

Answer:

hope it helps you dear