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How can I study the convergence of this series $$\sum\cos^n \frac{1}{n^\alpha}$$ depending on $\alpha >0$ ?

I found that $$\cos^n \frac{1}{n^\alpha} \sim \exp\left(\frac{-1}{2n^{2\alpha -1}}\right)$$

The case $2\alpha - 1 \geq 0$ can then be treated, as the series diverges. But how can I exploit that same similar to treat the other case?

Thanks

rtybase
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s89ne
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2 Answers2

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Using two facts, this one and this one, we have $$1-\frac{x^2}{2}\leq \cos{x}\leq e^{-\frac{x^2}{2}}, x \in \left[0,\frac{\pi}{2}\right]$$ or, because $\frac{1}{n^{\alpha}}$ will be close to $0$ for suficiently large $n$ $$1-\frac{1}{2n^{2\alpha}}\leq \cos{\frac{1}{n^{\alpha}}}\leq \frac{1}{e^{\frac{1}{2n^{2\alpha}}}}$$ and, applying Bernoulli's inequality $$1-\frac{n}{2n^{2\alpha}}\leq \left(1-\frac{1}{2n^{2\alpha}}\right)^n\leq \cos^n{\frac{1}{n^{\alpha}}}\leq \frac{1}{e^{\frac{n}{2n^{2\alpha}}}}$$

Thus, for $2\alpha-1\geq0$ $$\lim\limits_{n\to\infty}\cos^n{\frac{1}{n^{\alpha}}} \ne 0$$ and the series doesn't converge.

For $2\alpha-1<0$ or $0<1-2\alpha$ and $$0<\cos^n{\frac{1}{n^{\alpha}}}\leq \frac{1}{e^{\frac{n}{2n^{2\alpha}}}}= \left(\frac{1}{e}\right)^{\frac{n^{(1-2\alpha)}}{2}}$$

Using this limit (more details here) $$\lim\limits_{n\to\infty}\frac{n^{1-2\alpha}}{\ln{n}}=\infty$$ for suficiently large $n$ we will have $$n^{1-2\alpha}>4\ln{n}=\ln{n^4} \Rightarrow \\ e^{\frac{n^{1-2\alpha}}{2}}>e^{\frac{\ln{n^4}}{2}}=n^2 \Rightarrow \\ \left(\frac{1}{e}\right)^{\frac{n^{1-2\alpha}}{2}}<\frac{1}{n^2}$$ and by comparison test, the series converges.

rtybase
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Note that the expression we are dealing with now is $$\sum_{n=1}^\infty \exp(-n^a/2)$$ with $0<a<1$. Now we can use the integral test and a substitution $u^{1/a}=x$, $\frac{u^{1/a-1}}{a}du=dx$ $$\int_1^\infty \exp(-x^a/2)dx = \frac{1}{a}\int_1^\infty \exp(-u/2)u^{1/a-1}du$$ And the last integral converges via the ratio test.

Isaac Browne
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