Infinite series of tan(1/n)

[Kaura]

Question
_∞
∑ tan(1/n)
^{n = 1}
Does the infinite series diverge or converge?

Equations
If lim_{n → ∞} ≠ 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

 

[Mark44]

Question
_∞
∑ tan(1/n)
^{n = 1}
Does the infinite series diverge or converge?

Equations
If lim_{n → ∞} ≠ 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

What other tests do you know about?

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

 

[Kaura]

What other tests do you know about?

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

Wait I think I got it

Using the Limit Comparison Test with the infinite series of 1/n would give lim_{n → ∞} 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 lim_{n → ∞} 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

 

[Mark44]

Wait I think I got it

Using the Limit Comparison Test with the infinite series of 1/n would give lim_{n → ∞} 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 lim_{n → ∞} 1/cos(1/n) = 1/cos(0) = 1 so the limit = 1

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

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

Overall, looks good.

 

[Ray Vickson]

Wait I think I got it

Using the Limit Comparison Test with the infinite series of 1/n would give lim_{n → ∞} 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 lim_{n → ∞} 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

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 $\tan(1/n) > 1/(4n)$.