AP Calculus BC

Does the Sum of 1/n^n Converge? Yes

Q: Does the sum of 1/n^n converge?

Yes, by the root test with L = 0.

The sum of 1 over n to the n converges. The root test is the natural tool because the entire term is an nth power: taking the nth root leaves 1 over n, which tends to 0. A root limit below 1 proves convergence.

\sum_{n=1}^{\infty}\frac{1}{n^{n}}

Converges

Settled by the root test.

The root test

When the whole term is raised to the nth power, the root test strips the exponent away instead of comparing consecutive terms.

\sqrt[n]{\left|a_{n}\right|} = \sqrt[n]{\frac{1}{n^{n}}} = \frac{1}{n} \longrightarrow 0 < 1

The thresholds are identical to the ratio test: below $1$ converges, above $1$ diverges, exactly $1$ is inconclusive.

Which of the two to reach for

Factorials suit the ratio test, because consecutive factorials cancel. Whole-term nth powers suit the root test, because the root cancels the exponent. Both give the same verdict whenever both apply.

The mistakes students make

Confusing $n^{n}$ with $n^{2}$ or with $2^{n}$ . It grows faster than either, which is why this converges so fast.
Using the root test on a factorial. The nth root of $n!$ has no clean form; that is the ratio test's job.

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 1/n^n converge?

Yes, by the root test with $L = 0$ .

When should I use the root test over the ratio test?

When the entire term is an nth power. Factorials point to the ratio test instead.