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Discuss the convergence of the given series (Please apply any test to show convergence or divergence):

$$\sum_{k=1}^{\infty} \tan\left(\frac 1k\right)$$

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  • $\begingroup$ That’s homework, right? $\endgroup$ – k.stm Apr 1 '14 at 6:16
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Recall the Limit Comparison Test.

The Limit Comparison Test. Suppose $a_k,b_k>0$ for all $k$. If $\displaystyle\lim_{k\to\infty}\frac{a_k}{b_k}=c$ where $0<c<\infty$, then either both $\sum a_k$ and $\sum b_k$ converge or both diverge.

We can use the Limit Comparison Test to show that $\sum \tan(1/k)$ diverges.

Let $a_k=\tan(1/k)$ and $b_k=1/k$. Then $$ \lim_{k\to\infty}\frac{a_k}{b_k} = \lim_{k\to\infty}\frac{\tan(1/k)}{1/k} = \lim_{t\to0}\frac{\tan(t)}{t} \stackrel{\text{LH}}{=} \lim_{t\to0}\frac{\sec^2(t)}{1} = \sec^2(0) =1 $$ By the Limit Comparison Test, either both $\sum a_k$ and $\sum b_k$ converge or both diverge. Since $\sum b_k$ diverges, it follows that $\sum a_k$ diverges.

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  • $\begingroup$ nitpick: Isn't $\lim_{k\to\infty} \frac{\tan(\frac1{k})}{\frac1{k}} = \lim_{t\to0} \tan(t)/t$ valid and shorter? $\endgroup$ – Guy Apr 1 '14 at 6:50
  • $\begingroup$ @Sabyasachi I guess so. $\endgroup$ – Brian Fitzpatrick Apr 1 '14 at 6:52
  • $\begingroup$ Thanks Brian. That helped. Thank you so much. $\endgroup$ – Fahim Apr 6 '14 at 11:27
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Hint: For small positive $x$, we have $\tan x\gt x$.

Remark: We can prove this by noting that $\tan 0=0$ and that $\tan x-x$ is an increasing function on $(0,\pi/2)$.

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