Hi,(adsbygoogle = window.adsbygoogle || []).push({});

I have this series [tex]\sum_{k=1}^{\infty}\frac{\arcsin{\displaystyle\frac{1}{k}}}{k^2}[/tex]

I intended to prove this by Direct Comparison Test using the inequality:

[tex]\frac{\arcsin{\displaystyle\frac{1}{k}}}{k^2} \leq \frac{1}{k^2}[/tex]

Can I justify this inequality by saying that [tex]\arcsin{\frac{1}{x}}[/tex] is a decreasing function?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Series question

Loading...

Similar Threads - Series question | Date |
---|---|

I Question about Fourier series | Sep 19, 2016 |

I Is this even possible? Question about Fourier Series... | May 24, 2016 |

Taylor Series (Derivative question) | Dec 9, 2015 |

Alternating series question | Mar 18, 2015 |

Taylor Series Question | Jan 7, 2015 |

**Physics Forums - The Fusion of Science and Community**