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

1 Answer

Nov 16, 2016

#7#

Explanation:

#(x^(n+1)-1)/(x-1)=1+x+x^2+cdots+x^n# then

#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#

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