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Proving that the sum of $\sum_{n=9}^{\infty}\frac{1}{\sqrt(N)}= -\frac{1}{\sqrt(3)}$

Hi, I am trying to proving the sum above where $N$ is all the odd composites , any hint please ?

Brian Moehring
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Shopify lover
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    Do you really believe that the sum of a series of positive numbers can be negative? – José Carlos Santos Sep 27 '19 at 10:40
  • Yes, we already know that $1+2+3+4+5+...= \frac{-1}{12}$, here the wikipedia page https://fr.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF – Shopify lover Sep 27 '19 at 10:43
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    @PLAK-THEME We do not know that. That's not an equality. See, for instance, this post about the issue. That being said, you could ask us about whether your series is related to $-\frac1{\sqrt 3}$ the same way that $\sum n$ is related to $-\frac1{12}$. – Arthur Sep 27 '19 at 10:43
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    @PLAK-THEME $1+2+\cdots = -\frac{1}{12}$ is not true. – 5xum Sep 27 '19 at 10:44
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    @PLAK-THEME: No, $1+2+3+...=\infty $. – Surb Sep 27 '19 at 10:44
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    What exactly do you mean by "sum". The Wikipedia article you quote clearly states that taking the usual meaning the series diverges, but that some non-standard approaches can give a a "sum". Which approach are you interested in? Have you tried applying it to your $1/\sqrt{N}$ example? – almagest Sep 27 '19 at 10:51
  • Since you keep referencing the wikipedia article for the divergent sum $1+2+3+\cdots,$ which references zeta function regularization and ramanujan summation, I've changed the tag to match that. That said, I'm a little concerned that you're mistaking these summation methods as equivalent to the standard methods even after several people have tried to point out this flaw in your question to you. – Brian Moehring Sep 27 '19 at 11:26
  • I guess the question should be rephrased as: provided that $f(s)=\sum_{\substack{n\text{ odd}\n\text{ composite}}}\frac{1}{n^s}$ defines an analytic function over $\text{Re}(s)>1$, what is the value at $s=\frac{1}{2}$ of the analytic continuation of the previous series? – Jack D'Aurizio Sep 27 '19 at 16:54

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Since $\frac{1}{\sqrt{n}} > \frac1n$, and $$\sum_{n=9}^\infty \frac1n$$ diverges, it follows that the series $$\sum_{n=9}^\infty\frac{1}{\sqrt N}$$ also diverges.

5xum
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  • Intuitively the sum $\sum_{n=9}^\infty\frac{1}{\sqrt N}$ diverges as you mentioned, but i'm trying to use the same process as here https://fr.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF to prove that this converges to $\frac{-1}{\sqrt(3)}$ – Shopify lover Sep 27 '19 at 10:48
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    @PLAK-THEME There is no process that can be used to prove that a divergent sum converges. Divergent sums do not converge. – 5xum Sep 27 '19 at 10:55
  • ok got it thanks – Shopify lover Sep 27 '19 at 11:17