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Curious whether the partial sum of every convergent series can be stated as a closed form expression.

I saw this previous question, but they use the terminology "closed form value" and I'm not sure I understand what they mean.

By convergent series I mean an infinite sum of some expression where the partial sum approaches a single value in the limit.

An example is the sum $\sum_{n=1}^k 2 ^{-n}$ which can be written closed form as $2^{-k}(2^{-k} - 1)$

Is this possible for all convergent series?

If not, what is a counter example or how might you disprove it? Feel free to explain like I'm 5

Connor McCormick
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  • When $k=1,$ your example sum is just $2^{-1}=1/2.$ But your formula gives $-1/4$ when $k=1.$ – coffeemath Mar 08 '22 at 00:12
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    You can pretty much ignore the first sentence of Matt Samuel's answer if it helps. Basically, Matt's answer is a countability/uncountability argument (look up "Cantor's diagonal argument" for more details about what this means): every convergent series is associated with a real number, and there are more real numbers than there are ways to define/compute a number. The former is "uncountable", while the latter is "countable". – Theo Bendit Mar 08 '22 at 00:18
  • Take $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$. It's hard to imagine at least there is a closed form that converges to something about $\pi$. The point here is often the limit is much easier to compute than partial sums. – Just a user Mar 08 '22 at 00:24
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    Choose any uncomputable real number , say , $C$. Then for the series $C+\frac{C}{2}+\cdots+\frac{C}{2^n}$ (which converges to $2C$ for $n\to\infty$) there cannot be a closed form formula for the partial sum because inserting $n=1$ would give us a formula for $C$, hence $C$ would be computable. – Peter Mar 08 '22 at 13:04
  • @Peter of course you won’t capture the value $C$ in the closed form, but as n-> infinity you could. I’m just wondering if the partial sum of every infinite convergent series can be stated as a closed form expression. This doesn’t invalide that. If you give yourself C to start with, then why can’t my closed form also use C? – Connor McCormick Mar 08 '22 at 17:52
  • @TheoBendit doesn’t this though imply that convergent series can yield any range of reals, while a closed form expression cannot? Can’t my closed form itself contain a very specific real number which helps it to yield any possible real? – Connor McCormick Mar 08 '22 at 17:56
  • @Justauser agree it’s hard to imagine. I’m wondering if there’s a more rigorous way to capture that – Connor McCormick Mar 08 '22 at 17:57

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Not really. What is a "closed-form expression"? For example, if you start with polynomials and form all functions from them using $+,-,\times, /$ and $n$th roots (with any real number $n$), and composition of functions, and functional inverses, then after finitely many moves you still cannot get a function that equals $e^x$ for every $x\in [0,1].$ If you add $e^x$ into the mix and make more moves, there will be other analytic functions (power series) left out.

Another example: For $S\subseteq \Bbb N ,$ let $f(S)=\sum_{n\in S}2^{-n}.$ I don't expect to see a "closed-form expression" for $f$.

DanielWainfleet
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