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This is an alternating series because of $\sin{k}$. But if there is no $\pi/2$ in the argument for $\sin$, can I still replace $\sin{k}$ with $(-1)^k$? I don't think so.

Here is what I tried instead; the triangle inequality gives

$$\left|\sum_{k=1}^{\infty}\frac{\sin{(k)}\tan (1/k)}{\sqrt{k}}\right|\leq\sum_{k=1}^{\infty}\left|\frac{\sin{(k)}\tan (1/k)}{\sqrt{k}}\right|=\sum_{k=1}^{\infty}\frac{|\sin{(k)}|\tan (1/k)}{\sqrt{k}}.$$

I dont really know how to proceed and if this even makes sense because the triangle inequality applied like this requires that the series is absolutely convergent, and I've not shown that.

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    $\begingroup$ For an alternating series, successive terms should have opposite signs always, not just quite often. $\endgroup$ – Hagen von Eitzen Jan 1 '18 at 14:14
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    $\begingroup$ This is not alternating. Alternating needs to switch between positive and negative every term. $\endgroup$ – Michael Burr Jan 1 '18 at 14:15
  • $\begingroup$ You can apply the absolute value like this. The only possible problem is if the series on the right diverges, it doesn't tell you anything about the convergence of the left. $\endgroup$ – Michael Burr Jan 1 '18 at 14:18
  • $\begingroup$ Hints: what is the largest that $|\sin(k)|$ can be? Can you relate $\tan(x)$ to $x$ using an inequality for small $x$? $\endgroup$ – Michael Burr Jan 1 '18 at 14:20
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It ought to converge absolutely. Note that $\tan(1/k)\sim 1/k$ for large $k$.

So you have $$ \sum_{k=1}^\infty \left|\frac{\sin(k)}{\sqrt{k}}\tan(1/k)\right|=\sum_{k=1}^\infty \frac{|\sin(k)|}{\sqrt{k}}\tan(1/k)\leq\sum_{k=1}^\infty\frac{1}{\sqrt{k}}\tan(1/k)\sim\sum_{k=1}^\infty k^{-3/2} $$ convergent by the p test.

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  • $\begingroup$ Thanks man. How does one in words say $\sim$ ? Does this mean "asymptotically equal to"? $\endgroup$ – Parseval Jan 1 '18 at 14:21
  • $\begingroup$ yes that would be fine $\endgroup$ – qbert Jan 1 '18 at 14:23
  • $\begingroup$ But doesn't the application of the triangle inequality require the series to be absolutely convergent? I've not show that yet but still using it. $\endgroup$ – Parseval Jan 1 '18 at 14:24
  • $\begingroup$ yes, in fact you have shown that the series is absolutely convergent $\endgroup$ – qbert Jan 1 '18 at 14:24
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    $\begingroup$ we at no point assumed it was absolutely convergent. We bounded the absolute value of the terms in the series by something convergent. If you like names, this shows that the $\sum_n |a_n|$ converges by the comparison test $\endgroup$ – qbert Jan 1 '18 at 14:26
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Hint: We have $|\sin x|\le 1$, and for small $x$ we have $\tan x\approx x$.

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  • $\begingroup$ ugh, beat me to it $\endgroup$ – qbert Jan 1 '18 at 14:17
  • $\begingroup$ Damn, it's ridicolous how simple you guys make it. Thanks! $\endgroup$ – Parseval Jan 1 '18 at 14:20
  • $\begingroup$ if you see a function with a known taylor series with an argument getting small you should first "see" it's taylor expansion $\endgroup$ – qbert Jan 1 '18 at 14:24
  • $\begingroup$ What do you mean? $\endgroup$ – Parseval Jan 1 '18 at 14:24

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