?How do you find the sum of the infinite geometric series $\frac{6^{n}}{7^{n}}$ $6^n/7^n$ ?

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Answer 1 · 2016-11-16T10:12:57

$7$ $7$

$\frac{x^{n + 1} - 1}{x - 1} = 1 + x + x^{2} + \dots + x^{n}$ $(x^(n+1)-1)/(x-1)=1+x+x^2+cdots+x^n$ then

$n \sum k = 0 {(\frac{6}{7})}^{k} = \frac{{(\frac{6}{7})}^{n + 1} - 1}{(\frac{6}{7}) - 1}$ $sum_(k=0)^n (6/7)^k = ((6/7)^(n+1)-1)/((6/7)-1)$

Now,

$lim_(n->oo)sum_(k=0)^n (6/7)^k=lim_(n->oo) ((6/7)^(n+1)-1)/((6/7)-1)=1/(1-6/7)$ because

$lim_(n->oo)(6/7)^n=0$ so finally

$sum_(k=0)^oo (6/7)^k = 7$

?How do you find the sum of the infinite geometric series 6^n/7^n6n7n 6^n/7^n?