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Consider the periodic Dirichlet series that has this iterative definition:

$$\text{a1}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{6}}+...$$

$$\text{a2}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1}{\sqrt{6}}+...$$

$$\text{a3}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1+a2}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1+a2}{\sqrt{6}}+...$$

$$\text{a4}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1+a2+a3}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1+a2+a3}{\sqrt{6}}+...$$

$$...$$

continuing this iteration, will it converge to zero?

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2 Answers 2

up vote 3 down vote accepted

If I understand correctly, $a_1 = \sum_{j=0}^\infty (-1)^j \left((3j+1)^{-1/2} - (3j+2)^{-1/2} - 2 (3j+3)^{-1/2}\right)$, and $a_{j+1} = a_j + a_j b = a_j (1+b)$ where $b = \sum_{j=1}^\infty (-1)^j (3j)^{-1/2}$. Thus $a_n = a_1 (1+b)^{n-1}$. Now $1+b \approx .6507616$, so $|1+b| < 1$ so the sequence does converge to $0$.

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The original series is $$ \begin{eqnarray} a_1 &=& +\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}} - \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{5}}+\frac{2}{\sqrt{6}} + \dots \\ &=& \sum_{n=1}^{\infty}\frac{a_n}{\sqrt{n}}, \end{eqnarray} $$ where $a_n=2\cos\left(\frac{\pi}{3}n\right)=e^{i\pi n/3}+e^{-i\pi n/3}$. The successive iterates are $$ a_{k+1}=a_{k} + a_{k}\left(-\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{6}} - \frac{1}{\sqrt{9}} + \ldots\right) = a_{k}\left(1 + \frac{1}{\sqrt{3}}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}\right); $$ i.e., each is equal to the previous value times a fixed constant factor. The constant factor is $$ 1 + \frac{1}{\sqrt{3}}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}=1+\frac{(\sqrt{2}-1)\zeta(\frac{1}{2})}{\sqrt{3}}=0.65076...,$$ which is less than $1$, so the iterates converge to zero.

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