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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