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I'm trying to find out whether $\sum _{n=0}^{\infty }\left(\cos^n\left(\frac{1}{\sqrt{n}}\right)-\frac{1}{\sqrt{e}}\right)$ converges or not. I've tried with taylor series but it doesn't lead me anywhere except with the fact that $\lim_{n \to \infty}\cos^n\left(\frac{1}{\sqrt{n}}\right)-\frac{1}{\sqrt{e}}=0$ and therefore it has "a chance" to converge.

Any hint?

Sewer Keeper
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sky1099
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  • Hint:

    $$ \cos^n\left(\frac{1}{\sqrt n}\right) = \mathrm e ^{n \ln\cos\left(\frac{1}{\sqrt n}\right) } = \mathrm e ^{n \ln\left(1+\cos\left(\frac{1}{\sqrt n}\right) -1\right)} $$

     – Sewer Keeper May 02 '21 at 21:14
  • $\cos^n x = (\cos x)^n$, not the composition of $n$ cosines? – GEdgar May 02 '21 at 21:15
  • @GEdgar yes it's $(\cos x)^n$ – sky1099 May 02 '21 at 21:16
  • @SewerKeeper I did something similar, done the taylor series expansion but i end up stuck with $\exp(-\frac{1}{2}+\frac{1}{4n} + o(\frac{1}{\sqrt{n}}))$ and the thing inside $\exp$ doesn't tends to $0$ so I can't continue with taylor series? – sky1099 May 02 '21 at 21:18

3 Answers3

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The $n$-th term is asymptotic to $$-\frac{1}{12 \sqrt{e} n}$$ so, alas, the series diverges.

Igor Rivin
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hint

$$\cos(\frac{1}{\sqrt{n}})=1-\frac{1}{2n}+\frac{1}{24n^2}+o(\frac{1}{n^2})$$

$$\ln(1+\cos(\frac{1}{\sqrt{n}})-1)=$$

$$-\frac{1}{2n}-\frac{1}{12n^2}+o(1/n^2)$$

thus

$$\cos^n(\frac{1}{\sqrt{n}})=e^{n\ln(\cos(\frac{1}{\sqrt{n}}))}$$

$$=e^{-\frac 12}e^{-\frac 1n(\frac{1}{12}+o(1))}$$ $$=\frac{1}{\sqrt{e}}(1-\frac{1}{12n}(1+o(1))$$

hamam_Abdallah
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Note that $$ \cos\frac{1}{\sqrt{n}}\sim1-\frac{1}{2n}+\frac{1}{24n^2}+\ldots, $$ so $$ \log\cos\frac{1}{\sqrt{n}}= \log\left(1-\left(\frac{1}{2n}-\frac{1}{24n^2}+\ldots\right)\right)\\=-\left(\frac{1}{2n}-\frac{1}{24n^2}+\ldots\right)-\frac{1}{2}\left(\frac{1}{2n}+\ldots\right)^2+\ldots\\=-\frac{1}{2n}-\frac{1}{12n^2}+\ldots. $$ So $$ \cos^n\frac{1}{\sqrt{n}}=\exp\left(n\log \cos\frac{1}{\sqrt{n}}\right) \\ = \exp\left(-\frac{1}{2}-\frac{1}{12n}+\ldots\right) \\ =\frac{1}{\sqrt{e}}\exp\left(-\frac{1}{12n}+\ldots\right)\\ =\frac{1}{\sqrt{e}}-\frac{1}{12n\sqrt{e}}+O(n^{-2}). $$ Hence the terms are asymptotic to $-\frac{1}{12n\sqrt{e}}$, and the sum diverges.

mjqxxxx
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