Convergence / divergence examination $\sum _{k=1}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{k}\left(a\right)}{k}$

kramberol
2022-07-10
Answered

Convergence / divergence examination $\sum _{k=1}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{k}\left(a\right)}{k}$

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talhekh

Answered 2022-07-11
Author has **15** answers

If $\mathrm{sin}(a)=0$, then the series obviously converges.

If $\mathrm{sin}(a)=1$, then the series rewrites $\sum \frac{1}{k}$ which diverges.

If $\mathrm{sin}(a)=-1$, then the series rewrites $\sum \frac{(-1{)}^{k}}{k}$ which converges by Alternating Test.

If $|\mathrm{sin}(a)|<1$ and $\mathrm{sin}(a)\ne 0$, then one has

$\left|\frac{{\mathrm{sin}}^{k+1}(a)/(k+1)}{{\mathrm{sin}}^{k}(a)/k}\right|=|\mathrm{sin}(a)|\times \frac{k+1}{k}\u27f6|\mathrm{sin}(a)|$

Because $|\mathrm{sin}(a)|<1$, then by the ratio test, the series converges.

In conclusion, the series converges iff $\mathrm{sin}(a)\ne 1$, i.e.

$\overline{){\displaystyle \text{the series converges iff}a\ne {\displaystyle \frac{\pi}{2}}\text{mod}2\pi}}$

If $\mathrm{sin}(a)=1$, then the series rewrites $\sum \frac{1}{k}$ which diverges.

If $\mathrm{sin}(a)=-1$, then the series rewrites $\sum \frac{(-1{)}^{k}}{k}$ which converges by Alternating Test.

If $|\mathrm{sin}(a)|<1$ and $\mathrm{sin}(a)\ne 0$, then one has

$\left|\frac{{\mathrm{sin}}^{k+1}(a)/(k+1)}{{\mathrm{sin}}^{k}(a)/k}\right|=|\mathrm{sin}(a)|\times \frac{k+1}{k}\u27f6|\mathrm{sin}(a)|$

Because $|\mathrm{sin}(a)|<1$, then by the ratio test, the series converges.

In conclusion, the series converges iff $\mathrm{sin}(a)\ne 1$, i.e.

$\overline{){\displaystyle \text{the series converges iff}a\ne {\displaystyle \frac{\pi}{2}}\text{mod}2\pi}}$

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Consider the following series.

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The series is equivalent to the sum of two p-series. Find the value of p for each series.

Determine whether the series is convergent or divergent.

The series is equivalent to the sum of two p-series. Find the value of p for each series.

Determine whether the series is convergent or divergent.

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Consider, for all $n\in \mathbb{N}$ :

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Does the series