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I know that is the sequence of partial sums of a series is convergent, then the series is convergent. But let's say that the sequence of partial sums converges to the value $2$. Does that mean the sum of the series is $2$?

That doesn't make much sense to me because: as an example consider the series whose partial sums are given by the formula $s_n = 2 - 3(0.8)^n$. The limit as $n$ goes to infinity of that function is $2$. But wouldn't the sum of the series be much higher? Since: $s_1 + s_2 + s_3 + ... = 2-3(0.8) + 2-3(0.64)+2-3(0.512)+...$

Two is added with each new partial sum, so how can the sum of the series be two?

Sorry if this is a stupid question, I'm likely misunderstanding something fundamental about series. Any help is appreciated.

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A series converges by definition if the sequence of partial sums converges, and if that happens then the sum of the series is defined to be the limit of the sequence of partial sums.

I'll try to explain what is wrong with your example. You need to start from a sequence $(a_n)$ and define a corresponding sequence of partial sums $(S_n)$ by $S_n=\sum_{k=1}^n a_k$. Then $\sum_{k=1}^\infty a_k=\lim_{n\to\infty}\sum_{k=1}^n a_k=\lim_{n\to\infty} S_n$, if the limit exists.

Now in your example you already defined a sequence of partial sums $(s_n)$ and then started to look at its partial sums $s_1+...+s_n$. So you took partial sums of partial sums.

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  • $\begingroup$ Thank you for the response! Does the series $a_n$ in your example represent the sequence of terms of the series? And you define a sequence of partial sums by $S_n = $\sum_{k=1}^\infty a_k=\lim_{n\to\infty}\sum_{k=1}^n a_k=\lim_{n\to\infty} S_n$, but wouldn't that define a series and not a sequence? Sorry for not understanding, I appreciate the help! $\endgroup$ – James Ronald Apr 14 at 17:33
  • $\begingroup$ No, the elements $S_n$ are defined as finite sums $S_n=\sum_{k=1}^n a_k$. This is a sequence. The series is the infinite sum $\sum_{k=1}^\infty a_k$ which is defined to be the limit of the sequence $S_n$ if such a limit exists. And yes, the terms that we add are the elements of $\{a_k\}_{k=1}^\infty$. $\endgroup$ – Mark Apr 14 at 17:36
  • $\begingroup$ Ahh I see, sorry that was stupid of me. Everything makes sense now, thank you! $\endgroup$ – James Ronald Apr 14 at 17:40
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You are adding partial sums instead of finding the limit of partial sums.

That is why you are getting confused.

You have already taken care of the sigma by adding the terms of your series to find partial sums, so there is no need to add partial sums.

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