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hey I'm really confused on how to do infinite sums and how they work. this is one of the questions that i got. any help would be much appreciated: $$\sum_{n=1}^\infty x_n = \frac{p}{q}$$ $$n\in \mathbb{N}$$ $$p,q\in \mathbb{N}$$ for $$x_1=\frac{16}{35}$$ and $$x_{n+1}=\frac{-x_n^2}{1+x_n+x_n^2}$$ the original question asks what p + q is (both are relatively prime, forgot to add that) but i just wanna know how to evaluate the sum.
Again thanks.

Dominique
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ssecnirPmooDcitanuL
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  • Have you tried evaluating $x_2,x_3,\dotsc$ to see if there's a pattern to the terms or to the partial sum? – Gerry Myerson Oct 29 '24 at 12:05
  • Are you aware that $x_{n+1} = \frac{{x_n}^2-{x_n}^3}{{x_n}^3-1}$ (or something like that)? – Dominique Oct 29 '24 at 12:13
  • It appears that $p + q = 2 \times 16 + 35 = 67$. More generally, if $x_1 = a / b$ then $p / q = a / (a + b)$. – Jean Oct 29 '24 at 13:02
  • @Jean yes that is the correct answer. but how does it work tho? is there like a theorem im unaware of? – ssecnirPmooDcitanuL Oct 29 '24 at 13:36
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    So Jean is claiming that if $f(x)=-\frac{x^2}{1+x+x^2}$ and $x_{n+1}=f(x_n)$ then $\sum_{n=1}^{\infty}x_n=x_1/(1+x_1).$ (probably for $0<x_1<1)$ The special case where $x_1$ is rational does not seem important. This is a very surprizing result, and I long for its proof. – Letac Gérard Oct 29 '24 at 15:23
  • Excuse me as it's out of context, but still, can you please tell me from where have you got this problem? Actually I like to try out problems like these, but I don't generally encounter such problems. If you could tell me the resource(s) from where you got this problem, it would help me a lot in my mathematical career. Thanks! – FKcosθ Oct 29 '24 at 16:00
  • @LetacGérard - The first question to ask is if the sum even converges. Note that if $x_1 = 1$, then it would not, as every term but the first would be $-1$. I have not checked further, but as this is wildly divergent, I doubt other values close enough to $1$ would converge either. – Paul Sinclair Oct 29 '24 at 19:19
  • @FKcosθ it's a highschool level olympiad question from indonesia – ssecnirPmooDcitanuL Oct 29 '24 at 23:14
  • Have you tried my suggestion, ssecnir? Have you tried ANYTHING? Please meet us halfway. – Gerry Myerson Oct 31 '24 at 02:59
  • @GerryMyerson i have tried and there doesn't seem to be a pattern. i tried to use the infinite geometric sum thing and it didnt work out – ssecnirPmooDcitanuL Oct 31 '24 at 04:20
  • @Jean I would also be interested if you could elaborate. –  Oct 31 '24 at 11:29
  • @PaulSinclair If $x_1 = 1$ then $x_2 = -1/3 \neq -1$ and the sum does converge to $1/2$. However, if $x_1 = -1$ then $x$ is stationary and the sum diverges. – Jean Nov 01 '24 at 18:00
  • @ssecnirPmooDcitanuL you can show that $\forall |z| < 1, z - f(z) = \tilde{z} - f(\tilde{z})$ where $f$ is the sum for $x_1 = z$ and $\tilde{z} = -z/(1 + z)$. – Jean Nov 01 '24 at 18:05
  • @Jean - yes. I don't quite recall why I thought $1$ worked, – Paul Sinclair Nov 01 '24 at 21:39
  • @Jean for $z=\frac{16}{35}$, the equation you wrote becomes $f(\frac{16}{35})-f(-\frac{16}{51})=\frac{1376}{1785}$. But $f(-\frac{16}{51})$ is also unknown, isn’t it? So how does this help to find $f(\frac{16}{35})$? –  Nov 01 '24 at 21:58
  • To be honest, I really don’t think there is a formula to find the sum of the series for arbitrary $x_1$. The question just asked for $p+q$. So I’m pretty sure there must be some ridiculous trick that I can’t see for the moment. Like telescoping or something. –  Nov 01 '24 at 22:02
  • @abomasnow I believe otherwise, that the formula Letac wrote is indeed true, notably for all real $x > -1$. – Jean Nov 01 '24 at 22:58

1 Answers1

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Right, I think I found the “ridiculous trick” I was mentioning in the comment. Indeed telescoping is involved. I’ll just address the calculation of the sum of the series. Define a function $f$ by $f(x):=\frac{x}{x+1}$. Note that $$f(x_n)-f(x_{n+1})=\frac{x_n}{x_n+1}-\frac{-\frac{x_n^2}{1+x_n+x_n^2}}{-\frac{x_n^2}{1+x_n+x_n^2}+1}\\=\frac{x_n}{x_n+1}-\frac{-x_n^2}{1+x_n}=x_n.$$ Now the $N$-th partial sum becomes $$S_N=\sum_{n=1}^Nx_n=\sum_{n=1}^N\big(f(x_n)-f(x_{n+1})\big)=f(x_1)-f(x_{N+1}).$$ If the series converges, then $x_{N+1}\rightarrow 0$ as $N\rightarrow+\infty$, so also $f(x_{N+1})\rightarrow 0$ as $N \rightarrow +\infty$. So the sum of the series becomes $f(x_1)=\frac{x_1}{x_1+1}$.

If $x_1=\frac{16}{35}$, the sum of the series becomes $\frac{16}{51}$. So $p+q=16+51=67$.

  • By the way, here I just used this $f$ because I got inspired by the comments above. But I’d love to have a method to find it from scratch. I’m aware that there are methods to write, e.g., $x_n=y_n-y_{n-1}$ (Gosper’s algorithm), but given $x_n$ defined recursively, writing $x_n=f(x_n)-f(x_{n+1})$ is a different story. Perhaps I should ask a question specifically for this. –  Nov 02 '24 at 12:08
  • Very nice answer. Thank you very much. – Jean Nov 02 '24 at 18:11