Math, asked by rakshinisundar111, 10 days ago

Find the sum of series:

 - log_ntends  \: to  \:  \infty  \:  \frac{1}{6 }   +  \frac{1}{ {6}^{2} }  +  \frac{1}{ {6}^{3} }  + ........... \frac{1}{ {6}^{n} }

Answers

Answered by Anonymous
1

We have,

\displaystyle\lim_{n\to \infty}  \dfrac{1}{6} + \dfrac{1}{6^2}+\dfrac{1}{6^3} + \cdots + \dfrac{1}{6^n}

\displaystyle = \sum_{n=1}^\infty \dfrac{1}{6^n}

\displaystyle = \sum_{n=1}^\infty \left(\dfrac{1}{6}\right)^n

\displaystyle = \dfrac{\frac16}{1 - \frac16}

\displaystyle = \dfrac15

I've used formula of infinite GP to calculate the sum.

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