Find a nice form of $1+\frac23+\frac{6}{3^2}+\frac{10}{3^3}+\dots$ with sigma notation Question. Is it possible to find a nice form with sigma notation on the series: $1+\frac23+\frac{6}{3^2}+\frac{10}{3^3}+\dots$?
Attempt. I know this is rather a short question, specially if the answer to this is a no, but the original question is to find the numeric value of the sum, and what they did doesn't really look intuitive to me so I wanted to figure out if a sigma notation closed form was possible, since that would be what I'd have done. My attempt was: $$1+\sum^\infty_{n=0}\frac{4n+2}{3^{n+1}}$$ I don't even really know if putting out the $1$ is really possible or legit. In short terms, I'm looking for an intuitive approach with the first step to be: expressing that series (if possible) on sigma notation, so that later I can focus on finding the value. Still, don't know if it's possible to express it with sigma so yeah.
Edit. Still wondering if it's possible to find a numerical value to $$1+\sum^\infty_{n=0}\frac{4n+2}{3^{n+1}}$$
 A: You have a sequence with a sigma formula defining each term. $$1+\sum_0^\infty\frac{4n+2}{3^{n+1}}$$ You want a sum of those terms, which is a series. We can use your sequence to write the first few terms. For the moment, I will ignore the "1+" and try to remember to add it back in the end.
$$\frac{2}{3}, \frac{6}{9}, \frac{10}{27}, \frac{14}{81},\frac{18}{243},\frac{22}{729},\frac{26}{2187} \ldots$$
Next, we need to sum these and write the sequence of partiall sums.
$$\frac{2}{3}, \frac{12}{9}, \frac{46}{27}, \frac{152}{81}, \frac{1444}{729}, \frac{4358}{2187} \ldots$$ So far this is just arithmetic, but now comes the first bit of magic. The denominator of the partial sums is pretty much a given, as you figured it out for the recursion formula.  The denominator is $3^{n+1}$. After a bit of head scratching, the numerator is $2\cdot 3^{n+1}-2(n+2)$.  Now we just need the limit of the last term in the sum series.
$$
\lim_{n=0}^{\infty}\left[\frac{2\cdot3^{n+1}-2(n+2)}{3^{n+1}}\right]=\lim_{n=0}^{\infty}\left[\frac{2\cdot3^{n+1}}{3^{n+1}}\right]+\lim_{n=0}^{\infty}\left[\frac{-2(n+2)}{3^{n+1}}\right]=2
$$
Finally, remembering to add back your "1+" we get a limit of the summed series as $3$. That is, $3$ is the numerical value of the summation.
