I am shockingly terrible at determining whether or not infinite series converge or not... I'm stuck on the problem:

Does the series $\sum_{k=1}^{\infty} (\sin \frac{1}{k} - \arctan\frac{1}{k})$ converge?

I attempted solving it using Taylor approximations:

$$\sin \frac{1}{k} \approx \frac{1}{k} - \frac{1}{3!k^3} + O(\frac{1}{k^5})$$ $$\arctan \frac{1}{k} \approx \frac{1}{k} - \frac{1}{3k^3} + O(\frac{1}{k^5})$$ $$\sin \frac{1}{k} - \arctan \frac{1}{k} \approx \frac{1}{k} - \frac{1}{3!k^3} + O(\frac{1}{k^5}) - \left(\frac{1}{k} - \frac{1}{3k^3} + O(\frac{1}{k^5}) \right) \approx \frac{1}{6k^3} + O(\frac{1}{k^5})$$

$$\sum_{k=1}^{\infty} (\sin \frac{1}{k} - \arctan\frac{1}{k}) \approx \sum_{k=1}^{\infty} (\frac{1}{6k^3} + O(\frac{1}{k^5})) < \infty$$ So the series converges. $\square$

My questions are:

  1. Is my answer correct?

  2. Is there anything wrong with using Taylor approximations to determine convergence?

  3. Is there a rule of thumb for how to tackle questions which ask you to determine the convergence or divergence of series? If there is, please share - I could really use the help.

  • 2
    $\begingroup$ These are not $\approx$. Actually you could write $=$ as $k \to \infty$. If you used Taylor expansion right, then you were right. $\endgroup$
    – xbh
    Aug 19, 2018 at 19:09
  • 1
    $\begingroup$ Or you can use asymptotic equivalence: $\;\sin\dfrac1k-\arctan\dfrac1k\sim_\infty\dfrac6{6k^3}$. $\endgroup$
    – Bernard
    Aug 19, 2018 at 19:15

1 Answer 1


Your strategy does work. Why? Because the terms beyond $1/k^3$ are an arbitrarily small fraction of that term for sufficiently large $k$, so their contribution to large-$k$ terms are bounded on both sides by multiples of the $1/k^3$ part.

  • $\begingroup$ Ok, thank you. Do Taylor approx. always work or did I just get lucky here? $\endgroup$ Aug 19, 2018 at 19:17
  • 1
    $\begingroup$ @user9750060 If $f(k)=Ak^{-p}+o(k^{-p})$ for $p>1$, $\sum_{k\ge 1}f(k)$ will converge. $\endgroup$
    – J.G.
    Aug 19, 2018 at 19:30

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