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

Acyex
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    Without knowing what a general term is, we cannot give a unique answer. – Sal Jul 11 '21 at 19:27
  • What is the general term? Does this mean that putting it into the sigma notation is not possible? – Acyex Jul 11 '21 at 19:28
  • What is your problem with the form you got? What's the issue with putting the 1 out? – Sayan Dutta Jul 11 '21 at 19:29
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    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 '21 at 19:30
  • @SayanDutta I just didn't know if it was possible – Acyex Jul 11 '21 at 19:31
  • Oh, didn't know that, I was just given this. @Sal – Acyex Jul 11 '21 at 19:31
  • Is it now possible find a numerical value to this sum? – Acyex Jul 11 '21 at 19:32
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    @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. – Sayan Dutta Jul 11 '21 at 19:32
  • @Acyex it is possible to find the exact value, which I found to be 3 – fwd Jul 11 '21 at 20:07
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    The OEIS lists 174 different sequences beginning 1,2,6,10. And these are only the ones sufficiently well-known to be listed. – GEdgar Jul 11 '21 at 20:15
  • @fwd how did you do so? I'm currently stuck. Also I hope you are saying to have found a value of $1+\sum^\infty_{n=0}\frac{4n+2}{3^{n+1}}$ continuing from there, instead of just starting over and finding the value to $1+\frac23+\frac{6}{2^2}+\frac{10}{2^4}+\dots$ – Acyex Jul 11 '21 at 20:15
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    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 '21 at 20:18
  • It'd be amazing, please do thanks – Acyex Jul 11 '21 at 20:28
  • it does, thank you! – Acyex Jul 12 '21 at 13:33

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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.

Narlin
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