AP Calculus BC

Does the Sum of 2^n/n^n Converge? Yes

The series converges. The root test gives a limit of zero, which is as decisive as it gets. When every part of a term is already raised to the nth power, the nth root cancels all of it and the root test is the natural choice.

∑n=1∞2nnn\sum_{n=1}^{\infty}\frac{2^{n}}{n^{n}}

Converges

Settled by the root test.

The root test in one line

∣an∣n=2nnnn=2n⟶0\sqrt[n]{|a_n|} = \sqrt[n]{\frac{2^{n}}{n^{n}}} = \frac{2}{n} \longrightarrow 0

There is no approximation anywhere: the nth root of an nth power is exact. A limit of 0<10 < 1 proves absolute convergence, and 0 is as far from the inconclusive value of 1 as a limit can be.

When to reach for the root test

Use it when the whole term, or nearly all of it, sits under an nth power. Terms like (3n+12n+5)n\left(\frac{3n+1}{2n+5}\right)^{n} or 1(ln⁡n)n\frac{1}{(\ln n)^{n}} are built for it, and the ratio test on those is a mess.

The ratio test works here too, giving 2(nn+1)n⋅1n+1→02\left(\frac{n}{n+1}\right)^{n}\cdot\frac{1}{n+1} \to 0, but it takes several more steps. The two tests never disagree; they are inconclusive in exactly the same circumstances, at a limit of 1.

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

When is the root test better than the ratio test?

When the term is an nth power. The root cancels it exactly, while the ratio test has to expand and cancel. For factorials the ratio test is easier, because roots of factorials are awkward.

Can the root test and the ratio test disagree?

Never on a verdict. When both limits exist they are equal, and both are inconclusive at exactly 1. The root test does succeed in some cases where the ratio limit fails to exist.