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For which complex number $\text{z}$ does this series converge:

$$\sum_{\text{n}=1}^\infty\exp\left(2\pi\text{n}^2\text{z}i\right)$$

I used the ratio test, but it gives me $1$:

$$\lim_{\text{n}\to\infty}\left|\frac{\exp\left(2\pi\left(\text{n}+1\right)^2\text{z}i\right)}{\exp\left(2\pi\text{n}^2\text{z}i\right)}\right|=1$$

Sothe ratio test is not conclusive.

2 Answers 2

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You have made a mistake in your result of the ratio test.

Taking the ratio test as you did gives, as $n \to \infty$, $$ \left|\frac{\exp\left(2\pi\left(\text{n}+1\right)^2\text{z}i\right)}{\exp\left(2\pi\text{n}^2\text{z}i\right)}\right|=\left|\exp\left(4\pi nz i+2\pi z i\right)\right|=e^{\large -4\pi n b-2\pi b} \to \:? $$ with $z=a+ib$, $a,b \in \mathbb{R}$.

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    I got that the expression gives me: $$\exp\left(-2\pi\Im\left[\text{z}+2\text{n}\text{z}\right]\right)$$ and taking the limit as $\text{n}\to\infty$ it gives me $1$.. – 2017-01-08
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    @user403351 Please double check your result. You can see that the limit of the ratio goes to $0<1$. – 2017-01-08
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    When $\Im\left[\text{z}\right]>0$ the limit goes to $0$ – 2017-01-08
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    @user403351 Yes, and only in this case. – 2017-01-08
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    When $\text{z}=0$, we get: $$\sum_{\text{n}=1}^\infty1$$ can we say that, that equals $1$? – 2017-01-08
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    @user403351 No. By definition $\sum_{n=1}^\infty1=\lim_{N \to \infty}\sum_{n=1}^N1$ but $\sum_{n=1}^N1=N $ thus as $N \to \infty$, the given series is divergent in the case $\text{Im}z=0$. – 2017-01-08
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The $\;n\,$th root test gives:

$$\sqrt[n]{\left|e^{2\pi n^2zi}\right|}=\left|e^{2\pi nzi}\right|=e^{-2\pi n\,\text{Im}\,z}\xrightarrow[n\to\infty]{}0\iff \text{Im}\,z>0$$

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    For a series to converge, using the nth root test, that limit has to go to $0$, is that correct? – 2017-01-08
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    Nop! The above is only a sufficient condition, as the root test says if the supremum limit (of the absolute value of the general term) is **less than one** then the series converges *absolutely*. The above gives you a wide range (half the complex plane) of values for which the series is convergent...but it is not **necessarily** the only one. You should now check what happens when $\;\text{Im}\,z\ge0\;$ ... – 2017-01-08