# Infinite series of tan(1/n)

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1. Nov 2, 2016

### Kaura

Question

∑ tan(1/n)
n = 1
Does the infinite series diverge or converge?

Equations
If limn → ∞ ≠ 0 then the series is divergent

Attempt
I tried using the limit test with sin(1/n)/cos(1/n) as n approaches infinity which I solved as sin(0)/cos(0) = 0/1 = 0

This does not rule out anything and I cannot think of what else to try

The answer stated that the series is divergent so I just need to know what test to use to determine that

2. Nov 2, 2016

### Staff: Mentor

What other tests do you know about?

Also, in future posts, please do not delete the Homework Template.

3. Nov 2, 2016

### Kaura

Wait I think I got it

Using the Limit Comparison Test with the infinite series of 1/n would give limn → ∞ tan(1/n)/(1/n) = n tan(1/n) = n sin(1/n)/cos(1/n)
n sin(1/n) = sin(1/n)/(1/n) = 1 so the limit can be written limn → ∞ 1/cos(1/n) = 1/cos(0) = 1 so the limit = 1

Since the limit is larger ≥ 0 that means that both series tan(1/n) and 1/n must converge or diverge and since 1/n obviously diverges tan(1/n) also diverges

Please let me know if I made any mistakes and thank you

4. Nov 2, 2016

### Staff: Mentor

Better:
$\lim_{n \to \infty}\frac{\tan(1/n)}{1/n} = \lim_{n \to \infty} \frac{\sin(1/n)}{1/n} \frac 1 {\cos(1/n)} = \lim_{n \to \infty} \frac{\sin(1/n)}{1/n}\cdot \lim_{n \to \infty} \frac 1 {\cos(1/n)} = 1 \cdot 1 = 1$
Overall, looks good.

5. Nov 3, 2016

### Ray Vickson

You can use a comparison test: for small $x > 0$ we have $\sin x > x/2$ and $\cos x < 2$, so $\tan x > x/4$, hence $\tan(1/n) > 1/(4n)$.