5
$\begingroup$

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

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.

Advice?

$\endgroup$
  • 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
7
$\begingroup$

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 $$\frac{n^\frac{-1}{2}}{3}\leq \frac{n^\frac{-1}{2}}{2+\sin^2(n)}\leq \frac{n^\frac{-1}{2}}{2}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.