Question

Question: #10 [Class 11]

$\mathtt{Find \: the \: value \: of }\: \\ \mathfrak{ \sum_{n =1} ^{ \infin} \: \frac{1}{2n - 1} ( \frac{1}{ {9}^{n - 1} } + \frac{1}{ {9}^{2n - 1} } ) }$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\mathfrak{ \displaystyle\sum_{n =1} ^{ \infin} \: \frac{1}{2n - 1} \bigg( \frac{1}{ {9}^{n - 1} } + \frac{1}{ {9}^{2n - 1} } \bigg) }$

On simplifying the summation by substituting n = 1, 2, 3, ---

$\rm \: = \: \dfrac{1}{1} \bigg(\dfrac{1}{1} + \dfrac{1}{9}\bigg) + \dfrac{1}{3} \bigg(\dfrac{1}{9} + \dfrac{1}{ {9}^{2} }\bigg) + \dfrac{1}{5} \bigg(\dfrac{1}{ {9}^{2} } + \dfrac{1}{ {9}^{5} }\bigg) + - - - \infty$

can be re-arranged as

$\rm \: = \: \bigg(1 + \dfrac{1}{3.9} + \dfrac{1}{5. {9}^{2} } + - - \infty \bigg) + \bigg(\dfrac{1}{9} + \dfrac{1}{3. {9}^{3} } + \dfrac{1}{5. {9}^{5} } + - - \infty \bigg)$

can be further rewritten as

$\rm \: = \: \bigg(1 + \dfrac{1}{3. {3}^{2} } + \dfrac{1}{5. {3}^{4} } + - - \infty \bigg) + \bigg(\dfrac{1}{9} + \dfrac{1}{3. {9}^{3} } + \dfrac{1}{5. {9}^{5} } + - - \infty \bigg)$

can be further rewritten as

$\rm \: = \: 3\bigg(\dfrac{1}{3} + \dfrac{1}{3. {3}^{3} } + \dfrac{1}{5. {3}^{5} } + - - \infty \bigg) + \bigg(\dfrac{1}{9} + \dfrac{1}{3. {9}^{3} } + \dfrac{1}{5. {9}^{5} } + - - \infty \bigg)$

We know,

$\boxed{\tt{ x + \dfrac{ {x}^{3} }{3} + \dfrac{ {x}^{5} }{5} + - - - \infty = \frac{1}{2}log_{e}\bigg(\dfrac{1 + x}{1 - x} \bigg) \: }}$

So, using this result, we get

$\rm \: = \: 3 \times \dfrac{1}{2} log_{e}\bigg(\dfrac{1 + \dfrac{1}{3} }{1 - \dfrac{1}{3} } \bigg) + \dfrac{1}{2} log_{e}\bigg(\dfrac{1 + \dfrac{1}{9} }{1 - \dfrac{1}{9} } \bigg)$

$\rm \: = \: \dfrac{3}{2} log_{e}\bigg(\dfrac{\dfrac{3 + 1}{3} }{ \dfrac{3 - 1}{3} } \bigg) + \dfrac{1}{2} log_{e}\bigg(\dfrac{ \dfrac{9 + 1}{9} }{ \dfrac{9 - 1}{9} } \bigg)$

$\rm \: = \: \dfrac{3}{2} log_{e}\bigg(\dfrac{4}{2} \bigg) + \dfrac{1}{2} log_{e}\bigg({\dfrac{10}{8} } \bigg)$

$\rm \: = \: \dfrac{3}{2} log_{e}\bigg(2 \bigg) + \dfrac{1}{2} log_{e}\bigg({\dfrac{5}{4} } \bigg)$

We know,

$\boxed{\tt{ \: y \: logx \: = \: log {x}^{y} \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{1}{2} log_{e}\bigg( {2}^{3} \bigg) + \dfrac{1}{2} log_{e}\bigg({\dfrac{5}{4} } \bigg)$

$\rm \: = \: \dfrac{1}{2} log_{e}\bigg( 8 \bigg) + \dfrac{1}{2} log_{e}\bigg({\dfrac{5}{4} } \bigg)$

$\rm \: = \: \dfrac{1}{2}\bigg[log_{e}8 \: + \: log_{e}\bigg({\dfrac{5}{4} } \bigg)\bigg]$

We know,

$\boxed{\tt{ \: log_{e}x \: + \:log_{e}y \: = \: log_{e}xy \: }}$

So, using this we get

$\rm \: = \: \dfrac{1}{2}\bigg[ \: log _{e} \bigg({8 \times \dfrac{5}{4} } \bigg)\bigg]$

$\rm \: = \: \dfrac{1}{2} \: log _{e} 10$

Hence,

$\rm\implies \:\boxed{\tt{ \mathfrak{ \displaystyle\sum_{n =1} ^{ \infin} \: \frac{1}{2n - 1} \bigg( \frac{1}{ {9}^{n - 1} } + \frac{1}{ {9}^{2n - 1} } \bigg) } = \dfrac{1}{2}log_{e}10 \: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ \: log_{e}(1 + x) = x - \dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} - \dfrac{ {x}^{4} }{4} + - - - \infty \: }}$

$\boxed{\tt{ \: log_{e}(1 - x) = - x - \dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{3} }{3} - \dfrac{ {x}^{4} }{4} - - - - \infty \: }}$

$\boxed{\tt{ log_{e}2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + - - - \infty \: }}$

Answer 2

$\boxed{ \mathsf{Question-}}$

$\color{red}\mathtt{Find \: the \: value \: of }\: \\ \mathfrak{ \sum_{n =1} ^{ \infin} \: \frac{1}{2n - 1} ( \frac{1}{ {9}^{n - 1} } + \frac{1}{ {9}^{2n - 1} } ) }$

$\underline\mathrm{Solution:-}$

[Photo attached with the answer]

$\color{blue}\boxed{ \mathbf{Final \: Answer:}}$

$\mathsf{= \frac{1}{2} \: log_{e} ^{10} }$