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प्रश्न
\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\]
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उत्तर
\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . \frac{1}{3^n} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{1}{3}\left( 1 + \frac{1}{3} + \frac{1}{3^2} + . . . . . \frac{1}{3^{n - 1}} \right) \right]\]
\[ = \lim_{n \to \infty} \frac{\left[ \frac{1}{3} \left( 1 - \frac{1}{3^n} \right) \right]}{1 - \frac{1}{3}} \left[ a + ar + a r^2 + . . . . . . a r^{n - 1} = a\left( \frac{1 - r^n}{1 - r} \right) \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\frac{1}{3}\left( 1 - \frac{1}{3^n} \right)}{\frac{2}{3}} \right]\]
\[ = \lim_{n \to \infty} \frac{1}{2} \left[ \left( 1 - \frac{1}{3^n} \right) \right]\]
\[As n \to \infty , 3^n \to \infty , \frac{1}{3^n} \to 0\]
\[ = \frac{1}{2}\left( 1 - 0 \right)\]
\[ = \frac{1}{2}\]
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