1/5+1/15+1/45+1/81+... Converges or diverges. Sum or no sum

QUESTION POSTED AT 01/06/2020 - 04:06 PM

Answered by admin AT 01/06/2020 - 04:06 PM

Let's consider the n partial sum of the series:

S_n=\dfrac15+\dfrac1{15}+\dfrac1{45}+\dfrac1{81}+\cdots+\dfrac1{5\times3^{n-1}}
S_n=\dfrac15\left(1+\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}+\cdots+\dfrac1{3^{n-1}}\right)

Multiply both sides by \dfrac13, making sure to distribute \dfrac13 to each term in the sum on the right side:

\dfrac13S_n=\dfrac15\left(\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}+\dfrac1{3^5}+\cdots+\dfrac1{3^n}\right)

Now subtract this from S_n to get

S_n-\dfrac13S_n=\dfrac23S_n=\dfrac15\left(1-\dfrac1{3^n}\right)
\implies S_n=\dfrac3{10}\left(1-\dfrac1{3^n}\right)=\dfrac3{10}-\dfrac1{10\times3^{n-1}}

Now as n\to\infty, the second term converges to 0, leaving you with

\displaystyle\lim_{n\to\infty}S_n=\frac3{10}
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