AP Calculus BC

Does the Sum of arctan(n)/n Converge? No

The sum of arctan n over n diverges. Limit comparison with the harmonic series gives a ratio of pi over 2, a finite positive number, so the two series share a verdict, and the harmonic series diverges.

∑n=1∞arctan⁡nn\sum_{n=1}^{\infty}\frac{\arctan n}{n}

Diverges

Settled by the limit comparison test.

The numerator approaches pi over 2

Unlike a sine, arctan⁡n\arctan n does have a limit. It climbs towards a horizontal asymptote, getting arbitrarily close to π2\frac{\pi}{2} without ever reaching it.

lim⁡n→∞arctan⁡n=π2\lim_{n \to \infty}\arctan n = \frac{\pi}{2}

So for large nn each term is close to π2⋅1n\frac{\pi}{2}\cdot\frac{1}{n}, a fixed multiple of a harmonic term. A constant multiple cannot turn divergence into convergence.

Limit comparison with the harmonic series

Compare against bn=1nb_{n} = \frac{1}{n}. The ratio simplifies before any limit work is needed.

lim⁡n→∞arctan⁡nn1n=lim⁡n→∞arctan⁡n=π2\lim_{n \to \infty}\frac{\dfrac{\arctan n}{n}}{\dfrac{1}{n}} = \lim_{n \to \infty}\arctan n = \frac{\pi}{2}

Since π2\frac{\pi}{2} is finite and positive, both series do the same thing. The harmonic series diverges, so this one diverges.

A bounded numerator is not a rescue

Put the same numerator over n squared and the sum converges, because the p-series underneath converges. The numerator decides nothing on its own. The denominator does the deciding.

The mistakes students make

Every error below comes from expecting the bounded numerator to do more than it can.

  • Arguing that arctan⁡n\arctan n is bounded, so the series must converge. Boundedness only stops the numerator from making matters worse; ∑1n\sum \frac{1}{n} still diverges underneath.
  • Using arctan⁡n<π2\arctan n < \frac{\pi}{2} for a direct comparison and calling it convergent. That bounds the terms above by a divergent series, which settles nothing.
  • Applying the nth term test, finding arctan⁡nn→0\frac{\arctan n}{n} \to 0, and reporting convergence. That test can only prove divergence.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Does the sum of arctan(n)/n converge?

No. It diverges by limit comparison with the harmonic series, with ratio π2\frac{\pi}{2}.

Why does arctan(n)/n^2 converge but arctan(n)/n diverge?

The numerator is the same in both and is bounded in both. The denominator decides: ∑1n2\sum \frac{1}{n^{2}} converges and ∑1n\sum \frac{1}{n} diverges.

The terms go to 0, so why does it not converge?

Terms tending to 00 is necessary for convergence but never sufficient. The harmonic series is the standard counterexample, and this series is a multiple of it in the limit.