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Question
Find the sum of the following series to infinity:
`1/3+1/5^2 +1/3^3+1/5^4 + 1/3^5 + 1/56+ ...infty`
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Solution
Given series has sum S = `1/3+1/5^2 +1/3^3+1/5^4 + 1/3^5 + 1/56+ ...infty`
S = `(1/3+1/3^3+1/3^5...infty)+(1/5^2+1/5^4+1/5^6...infty)`
We know sum of a G.P. up to infinity is given by, S = `a/(1 – r)`.
Let S1 = `1/3+1/3^3+1/3^5...infty`
This is an infinite G.P. with first term(a) = 1/3 and common ratio(r) = 1/32 = 1/9.
So, S1 = `(1/3)/(1-1/9)`
= `3/8`
Let S2 = `1/5^2+1/5^4+1/5^6...infty`
This is an infinite G.P. with first term (a) = `1/5^2` and common ratio (r) = `1/5^2 = 1/25`.
So, S2 = `(1/25)/(1-1/25)`
= `1/24`
Now, required sum, S = S1 + S2
= `3/8+1/24`
= `10/24`
= `5/12`
Therefore, sum of the series to infinity is `5/12`.
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