# Find a nice form of $1+\frac23+\frac{6}{3^2}+\frac{10}{3^3}+\dots$ with sigma notation [duplicate]

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 $$1+\sum^\infty_{n=0}\frac{4n+2}{3^{n+1}}$$

• Without knowing what a general term is, we cannot give a unique answer.
– Sal
Jul 11, 2021 at 19:27
• In your post you conjectured a general term like $\frac{4n+2}{3^{n+1}}$, which is one possibility among many that generate the first four numbers in your question
– Sal
Jul 11, 2021 at 19:30
• @Acyex I think, it's absolutely fine. Also, $1,2,6,10,\dots$ doesn't follow any spwcific order (with the $1$ in front). So, don't think there's anything wrong with your form. Jul 11, 2021 at 19:32
• The OEIS lists 174 different sequences beginning 1,2,6,10. And these are only the ones sufficiently well-known to be listed. Jul 11, 2021 at 20:15
• I meant the sum $1 + \sum_{n=0}^{\infty} \frac{4n + 2}{3^{n+1}}$, and I found it using power series methods. I can write this is as an answer if you want.
– fwd
Jul 11, 2021 at 20:18

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.