SalmonPanna2022
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Lim n-> infinity [1/3 + 1/3² + 1/3³ + . . . .+ 1/3ⁿ]​

Sagot :

SalmonWorldTopper

Answer:

[tex]\large\underline{\sf{Solution-}}[/tex]

Given expression is

[tex]\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg][/tex]

Let we first evaluate

[tex]\rm :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } [/tex]

Its a Geometric progression with

[tex]\rm :\longmapsto\:a = \dfrac{1}{3} [/tex]

[tex]\rm :\longmapsto\:r = \dfrac{1}{3} [/tex]

[tex]\rm :\longmapsto\:n = n[/tex]

So, Sum of n terms of GP series is

[tex]\rm :\longmapsto\:S_n = \dfrac{a(1 - {r}^{n} )}{1 - r} [/tex]

[tex]\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{1 - \dfrac{1}{3} } \bigg][/tex]

[tex]\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{3 - 1}{3} } \bigg][/tex]

[tex]\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{2}{3} } \bigg][/tex]

[tex]\bf\implies \:S_n = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg][/tex]

Hence,

[tex]\bf :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg][/tex]

Therefore,

[tex] \purple{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]}[/tex]

[tex]\rm \:  =  \: \displaystyle\lim_{n \to \infty }\rm \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg][/tex]

[tex]\rm \:  =  \: \rm \dfrac{1}{2}\bigg[1 - 0 \bigg][/tex]

[tex]\rm \:  =  \: \rm \dfrac{1}{2}[/tex]

Hence,

[tex] \purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]} = \frac{1}{2}}}[/tex]

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Explore More

[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}[/tex]

[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}[/tex]

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[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}[/tex]

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