Question

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

Answer 1

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

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