Question

PLEASE SOLVE IT...

FIND THE LIMITING VALUE ...


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Answer 1

Solution

We have

$\displaystyle \lim _{ \tt{n \to \infty}} \tt\frac{ \bigg[1 + \dfrac{1}{2} + \dfrac{1}{ {2}^{2} } + ............... + \dfrac{1}{2^{n} } \bigg ]}{ \bigg[1 + \dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + ............... + \dfrac{1}{ {3}^{n} } \bigg]}$

Now Take

$\tt \to\bigg[1 + \dfrac{1}{2} + \dfrac{1}{ {2}^{2} } + ............... + \dfrac{1}{2^{n} } \bigg]$

So this Sequence is GP , We use sum of infinite GP is

$\tt \to S _{n} = \dfrac{1(1 - r ^{n} )}{1 - r}$

Using this formula we get

$\tt \to \: 1 \bigg(\dfrac{1 - \bigg(\dfrac{1}{2} \bigg)^{n + 1} }{1 - \dfrac{1}{2} } \bigg)$

$\tt \to \: \dfrac{ {2}^{n + 1} - 1}{ {2}^{n + 1} \times \dfrac{1}{2} } = \dfrac{ {2}^{n + 1} - 1}{ {2}^{n + 1} \times {2}^{ - 1} } = \dfrac{ {2}^{n + 1} - 1 }{ {2}^{n} }$

Now take

$\tt\to{\bigg[1 + \dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + ............... + \dfrac{1}{ {3}^{n} } \bigg]}$

So this Sequence is GP , We use sum of infinite GP is

$\tt \to S _{n} = \dfrac{1(1 - r ^{n} )}{1 - r}$

We get

$\tt \to \: \dfrac{1 \bigg(1 - \bigg( \dfrac{1}{3} \bigg)^{2n + 1} \bigg)}{1 - \dfrac{1}{3} }$

$\to \tt \: \dfrac{ {3}^{n + 1} - 1}{ {3}^{n + 1} \times \dfrac{2}{3} } = \dfrac{ {3}^{n + 1} - 1}{2 \times {3}^{n} }$

Now Put the Value

$\to \displaystyle \lim _{ \tt{n \to \infty}} \tt\dfrac{ \dfrac{ {2}^{n + 1} - 1 }{ {2}^{n} } }{ \dfrac{ {3}^{n + 1} - 1}{2 \times {3}^{n} } }$

$\to\displaystyle \lim _{ \tt{n \to \infty}} \tt\dfrac{2 - \dfrac{1}{ {2}^{n} } }{ \dfrac{3}{2} - \dfrac{1}{2 \times {3}^{n} } }$

$\to\tt\dfrac{2 - \dfrac{1}{ {2}^{ \infty } } }{ \dfrac{3}{2} - \dfrac{1}{2 \times {3}^{ \infty } } }$

$\to\tt\dfrac{2 - 0}{ \dfrac{3}{2} - 0 } = 2 \times \dfrac{2}{3} = \dfrac{4}{3}$

Answer is

$\to \: \dfrac{4}{3}$

Answer 2

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