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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?

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# Series question

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