$$\displaystyle\sum\limits_{n=1}^\infty \dfrac{n^{-1/2}}{2+\sin^2(n)}$$

I tried the divergence test but it approaches 0.

It's not a geometric series.

Doubt I could apply the integral test.

Not a p-series nor an alternating series.

Can't use root test.

I suppose I am left with either ratio test or a comparison test but I couldn't figure those out.


  • 4
    $\begingroup$ Hint: all terms are $\ge \frac{1}{3}\cdot\frac{1}{n^{1/2}}$. $\endgroup$ – André Nicolas Nov 8 '11 at 15:06

Hint: Try using the comparison test and comparing to the function $\frac{1}{\sqrt{n}}$. To apply this, notice that $2\leq 2+\sin^2(n)\leq 3$ so that $$\frac{n^\frac{-1}{2}}{3}\leq \frac{n^\frac{-1}{2}}{2+\sin^2(n)}\leq \frac{n^\frac{-1}{2}}{2}.$$


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