How do you find the sum of #1+1/5+1/7^2+1/7^3+1/7^9+...+1/7^n+...#?

1 Answer
Feb 2, 2017

#1+1/5+1/7^2+1/7^3+1/7^9+...+1/7^n+... = 1+1/5+1/49+1/343 +7/6-282475242/242121642 ~= 1.223323644#

Explanation:

The question is not very clear...in answering I assume that for #n>=9# the sum is on all values of #n#

We can write the sum as:

#1+1/5+1/49+1/343 + sum_(n=9)^oo 1/7^n#

Focus on the series: clearly we have:

#sum_(n=9)^oo 1/7^n = sum_(n=0)^oo 1/7^n - sum_(n=0)^8 1/7^n#

Thus at the second member we have the sum of a geometric series and the partial sum of a geometric series of ratio #1/7#:

#sum_(n=0)^oo 1/7^n = sum_(n=0)^oo (1/7)^n = 1/(1-1/7) = 7/6#

#sum_(n=0)^8 1/7^n = sum_(n=0)^8 (1/7)^n = (1-1/7)^9/(1-1/7) = 7/6(1-1/40353607) = 282475242/242121642#

So:

#1+1/5+1/7^2+1/7^3+1/7^9+...+1/7^n+... = 1+1/5+1/49+1/343 +7/6-282475242/242121642 ~= 1.223323644#