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Given $a, b \in \mathbb{R}$ with $b > 0$, is the series

\begin{equation} \sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| ) \end{equation}

convergent or divergent?

If we drop out the $n=0$ term and fold the remaining sum, the question can be equivalently asked for the series

\begin{equation} \sum_{n = 1}^{\infty} (\sqrt{(a+n)^2+b^2} + \sqrt{(a-n)^2+b^2} - 2n ). \end{equation}

Talley-rond
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3 Answers3

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My gut reaction is to say divergent, since the asymptotic behaviour (assume $n>0$, since the series is only twice that, plus a finite number of terms) is something like $$ \begin{align} \sqrt{n^2 + a^2 + 2an + b^2} - n &= n\left(\sqrt{1 + \frac{a^2}{n^2} + \frac{2a}{n} + \frac{b^2}{n^2}} - 1\right) \\ &= n\left(1 + \frac{a}{n} + o\left(\frac{1}{n}\right) - 1\right) \\ &= a + o(1) \end{align} $$

which seems to diverge for any $a \ne 0$.

(note that this assumes that $[\cdot]$ is just a normal set of brackets, if it's the floor function, then there are only a finite number of non-zero terms).

Guillermo Angeris
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  • What are you assuming about the order of the terms? I don't really understand what sum over $n\in \mathbb{Z}$ means ... is there some convention for that? Also are you the same Guillermo from PROMYS? I think we were there at the same time! :) – Zubin Mukerjee Aug 26 '17 at 02:57
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    @ZubinMukerjee I am indeed! (actually just messaged you on FB :) ) And the idea is that if the series diverges everywhere, the order of the terms is irrelevant, since each term will be positive and bounded from below for some large enough $n$. Note that the sign of each term is always positive, too! – Guillermo Angeris Aug 26 '17 at 02:58
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    No wait, you're totally right, order will matter here an infinite number of terms will be negative when summing over $n < 0$! @ZubinMukerjee so this isn't well-defined unless one is talking about only positive or only negative $n$. – Guillermo Angeris Aug 26 '17 at 03:03
  • This is true in the limit $n \rightarrow \infty$. But the $n \rightarrow - \infty$ contribution exactly cancels this. – Talley-rond Aug 26 '17 at 04:46
  • @Talley-rond Note that in the comments it's mentioned that the order of the sum matters, so this isn't well-defined since this is nowhere absolutely convergent. – Guillermo Angeris Aug 26 '17 at 21:13
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$$ \begin{align} \lim_{n \to \infty }\sqrt{a^2+2an+n^2+b^2}-n=\lim_{n \to \infty }\frac{\left ( a^2+2an+b^2 \right )}{\sqrt{a^2+2an+n^2+b^2}+n}=a \end{align} $$ evidently divergent for $$ \begin{align} a\neq 0 \end{align} $$

Tongho
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  • Yes, but: the sum is over all of $\mathbb{Z}$, and the limit at $-\infty$ is equal to $-a$. If you pair asymptotic terms up, they cancel each other in the large $n$ limit. I edited the question to make this a bit more clear. – Talley-rond Aug 26 '17 at 04:42
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Your two expressions are not equivalent. The sum is not uniformly convergent, so the order you sum the terms in can matter. The second has an explicit order of summing while the first does not and will cancel out the constant terms. Even the second one does not converge, however.

We have $\sqrt {(n+a)^2+b^2}-n \sim a+\frac {b^2}{2n}+o(n^{-1})$ as $n$ gets large, so $\sqrt {(n+a)^2+b^2}+\sqrt {(n-a)^2+b^2}-2n \sim\frac {b^2}{n}+o(n^{-1})$ and the sum diverges logarithmically

Ross Millikan
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  • In the last equation you wrote there is a little typo, one of the $n+a$ must be changed to $n-a$. But other than that, the argument, I think, makes sense. Thank you! – Talley-rond Aug 26 '17 at 16:03
  • You are right. Thanks. – Ross Millikan Aug 26 '17 at 17:10
  • Based on your answer, I rephrased the question here in such a way as to have the logarithmic tail removed. As it happens, after a more careful analysis of my problem, $that$ should have been the question I should have asked in the first place. – Talley-rond Aug 26 '17 at 20:37