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Here's the limit.

$$\lim_{n\to\infty}-\frac{4z}{1+4z^2}n+\sum_{j=1}^n\sqrt\frac{n}{j}\sin(z\log{\frac{n}{j}})$$

Can we prove that this doesn't converge for any $z$ where $\Im(z)\ne0$?

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  • $\begingroup$ The convergence In C even now is not clear or less obvious at all $\endgroup$ – zeraoulia rafik Jan 6 '17 at 22:07
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This is just a note.

There is a Riemann sum involved here: $$\lim_{n\rightarrow\infty}R_n(z):=\lim_{n\rightarrow\infty}\,\,\frac1n\sum_{j=1}^n\sqrt\frac{n}{j}\sin\left(z\log{\frac{n}{j}}\right) =-\int_0^1\frac{\sin(z\log x)}{\sqrt{x}}\,dx=\frac{2z}{1+4z^2};$$ valid (at least) for $z\in\mathbb{R}$. In view of this, the OP's question amounts to: how fast does the Riemann sums converge to the integral with respect to $n$; that is, $$R_n(z)-\int_0^1\frac{-\sin(z\log x)}{\sqrt{x}}\,dx=O(??) \tag1$$ Some numerical evidence suggests that (even for real $z\neq0$) the order of convergence is weaker than $\frac1n$, which means the OP's limit does not converge.

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The sum can be carried out in closed form in terms of harmonic numbers:

$$\frac{1}{n}\sum _{j=1}^n \sqrt{\frac{n}{j}} \sin \left(z \log \left(\frac{n}{j}\right)\right)=-\frac{i}{2 \sqrt{n}}\left(n^{i z} H_n^{\left(i z+\frac{1}{2}\right)}-n^{-i z} H_n^{\left(\frac{1}{2}-i z\right)}\right)$$

Now we can expand the harmonic numbers at $n=\infty$. The leading terms are:

$$\frac{4 z}{4 z^2+1}+\frac{1}{2} i \sqrt{\frac{1}{n}} \left(n^{-i z} \zeta \left(\frac{1}{2}-i z\right)-n^{i z} \zeta \left(\frac{1}{2}+i z\right)\right)$$

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