AP Calculus BC

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

Q: Why is the limit pi over 2?

Because \arctan n \to \pi/2 as n \to \infty: the arctangent function has a horizontal asymptote there. Taking the nth root of the nth power leaves exactly \arctan n.

Q: Would the ratio test work here?

It is far messier. The quotient is \frac{(\arctan(n+1))^{n+1}}{(\arctan n)^{n}}, which does not simplify cleanly. Terms built as nth powers belong to the root test.

The series diverges. The root test gives a limit of pi over 2, roughly 1.571, which is greater than one. Because arctangent settles just above one rather than below it, the terms eventually grow like a geometric series with that ratio.

\sum_{n=1}^{\infty}\left(\arctan n\right)^{n}

Diverges

Settled by the root test.

The root test does all the work

\sqrt[n]{|a_n|} = \arctan n \longrightarrow \frac{\pi}{2} \approx 1.5708

The nth root of an nth power is exact, so the limit is just the limit of $\arctan n$ , which is $\pi/2$ . Since $\pi/2 > 1$ , the root test proves divergence.

The whole verdict turns on $\pi/2$ being bigger than 1. Had the base settled below 1, the same computation would have proved convergence just as cleanly.

Why the terms are deceptive at first

Early terms are small: $\arctan 1 = \pi/4 \approx 0.785$ , so the first term is under 1, and $(\arctan 2)^{2} \approx 1.23$ . The base only crosses 1 once $\arctan n > 1$ , which happens at $n = 2$ .

After that the growth is relentless, since a base above 1 raised to the nth power runs away. The nth term test would also catch this, but only after you have worked out that the terms grow, which is essentially the same computation.

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

Why is the limit pi over 2?

Because $\arctan n \to \pi/2$ as $n \to \infty$ : the arctangent function has a horizontal asymptote there. Taking the nth root of the nth power leaves exactly $\arctan n$ .

Would the ratio test work here?

It is far messier. The quotient is $\frac{(\arctan(n+1))^{n+1}}{(\arctan n)^{n}}$ , which does not simplify cleanly. Terms built as nth powers belong to the root test.