How do you find the sum of the infinite series #Sigma2(1/10)^k# from k=1 to #oo#?

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Answer 1 · 2017-02-10T14:27:57

#sum_(n=1)^oo 2(1/10)^n = 2/9#

Start from the geometric series of ratio #r=1/10#:

#sum_(n=0)^oo (1/10)^n = 1/(1-1/10) = 10/9#

Extract from the sum the term for #n= 0#:

#1+ sum_(n=1)^oo (1/10)^n =10/9#

#sum_(n=1)^oo (1/10)^n =10/9-1 = 1/9#

Multiplying by #2# term by term:

#sum_(n=1)^oo 2(1/10)^n = 2 sum_(n=1)^oo (1/10)^n = 2/9#