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Here is the sum:

$\sum_{n=1}^{\infty}(-1)^n(\frac{1}{2n-1})\tan (\frac{1}{\sqrt{n}})$

Test for absolute convergence, conditional convergence, or divergence.

I used the alternating series test and found that it converges. How do I go about testing for absolute convergence? I tried comparing it to $\sum\frac{1}{n}$ with the Limit Comparison Test but the limit got really hairy when doing L'Hopital's rule. I tried comparing $\sum\tan(\frac{1}{\sqrt{n}})$ to $\sum\frac{1}{\sqrt{n}}$ but I don't know which is bigger. Any thoughts?

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  • $\begingroup$ After taking an absolute value and recognizing that $\tan(x) \approx x$ when $x \approx 0$, the summands are of order $(2n \sqrt{n})^{-1}$. What can you conclude now? $\endgroup$
    – user296602
    Commented Oct 11, 2017 at 3:28
  • $\begingroup$ This only works when n is small...what about when n is large? Are the summands still on the same order? $\endgroup$
    – VV6570
    Commented Oct 11, 2017 at 3:30
  • $\begingroup$ No, it works when $1/\sqrt{n}$ is small, which is exactly when $n$ is large. $\endgroup$
    – user296602
    Commented Oct 11, 2017 at 3:32
  • $\begingroup$ Oh my mistake, was looking at the wrong argument for tan(). Thanks! $\endgroup$
    – VV6570
    Commented Oct 11, 2017 at 3:37

2 Answers 2

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For small $x$, both $\sin(x)$ and $\tan(x)$ are close to (i.e., within a factor of 2 of) $x$, so that replacement can almost always be done.

And, of course, $\dfrac1{2n-1}\le \dfrac1{n}$.

Therefore the product is less than twice $\dfrac1{n^{3/2}}$, and the sum of this converges.

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Compare $a_n=\dfrac{1}{2n-1}\tan {\dfrac{1}{\sqrt n}}$

with $u_n=\dfrac{1}{n^{\frac{3}{2}}}$

Then we have $\lim\dfrac{a_n}{u_n}=\dfrac{\frac{1}{2n-1}\tan \dfrac{1}{\sqrt n}}{n^{\frac{3}{2}}}=\dfrac{\tan \frac{1}{\sqrt n}}{\frac{1}{\sqrt n}}\times \dfrac{1}{2-\frac{1}{n}}\to \frac{1}{2}(\neq 0)$ as $n\to \infty$

Since $u_n$ converges so the series $\sum a_n$ converges absolutely.

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