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Both the sum of $1+2+3+4+\cdots$ and the sum of $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots$ diverge. If both are paired together in one function, as seen above, can they amount to a number or can it even be calculated whatsoever?

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    $\begingroup$ Since $x^{\mathrm{sign}(x)}$ does not converge, much less to zero, how could the above series converge? $\endgroup$ – Thomas Andrews Jul 2 '15 at 20:53
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Let $f:\mathbb{N}\to\mathbb{N}$ be a function with $\lim_{n\to\infty}f(n)=\infty$. One can define

$$S_n(f)=\sum_{k=-f(n)}^n k^{\text{sign}(k)}=-\left({1\over f(n)}+\cdots+{1\over2}+{1\over1}\right)+1+2+\cdots+n$$

and then ask for $\lim_{n\to\infty}S_n(f)$. It's fairly clear you can tailor $f$ to get any limit you want. In particular, $f(n)=\lfloor e^{n(n+1)/2}\rfloor$ should lead to the Euler-Mascheroni constant, or something like it.

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That would diverge. Just take $(1-\frac{1}{2})+(2-\frac{1}{3})+(3-\frac{1}{4})+\cdots>\frac{1}{2}+\frac{1}{2}+\cdots$, which also diverges.

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It depends on what do you mean by: $$\sum\limits_{n = -\infty}^{\infty} n^{sign(n)}$$

Most people would say that:

$$\sum\limits_{n = -\infty}^{\infty} a_n=\sum\limits_{n = 0}^{-\infty}a_n+\sum\limits_{n = 1}^{\infty} a_n $$

where both converge. In this sense, it does not converge. If you want to take some type of principal value, it still doesn't converge. If you want to take a "principal value":

$$\sum\limits_{n = -\infty}^{\infty} n^{sign(n)}=\sum\limits_{n=1}^{\infty} n-\frac{1}{n}$$ Which doesn't converge either. Try taking the limits different ways. It shouldn't be hard to convince yourself that it won't converge.

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  • $\begingroup$ I don't count any fourier analyst to "most people" then... Technically, in analogy to improper integrals, $$\sum_{n=-\infty}^\infty a_n := \lim_{k\to\infty} \sum_{n=l_k}^{m_k} a_n$$ For any sequences such that $l_k \to -\infty$ and $m_k \to \infty$. Well-definedness of said thing can be worked out if $$\lim_{N\to\infty} \sum_{n=-N}^N |a_n| < \infty$$ This allows simplification in case of convergence to $$\sum_{n=-\infty}^\infty a_n = \lim_{N\to\infty} \sum_{n=-N}^N a_n$$ $\endgroup$ – AlexR Jul 2 '15 at 21:19

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