AP Calculus BC

Does the Sum of 1/(ln n)^n Converge? Yes

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

Yes, absolutely. The root test gives L = 0 < 1.

Q: Why does this series start at n = 2?

Because \ln 1 = 0, so the term at n = 1 would require division by zero.

Q: When should I pick the root test over the ratio test?

When the term is an nth power. The root test strips the exponent, while the ratio test leaves you comparing two different bases.

The sum of 1 over ln n to the n converges absolutely. The root test decides it: the nth root of the term is 1 over ln n, which tends to 0, and 0 is less than 1. The series starts at n equals 2 because ln 1 is 0.

\sum_{n=2}^{\infty}\frac{1}{\left(\ln n\right)^{n}}

Converges

Settled by the root test.

The whole term is an nth power

When the entire term is something raised to the power $n$ , the root test removes that exponent in a single step.

\sqrt[n]{\left|\frac{1}{(\ln n)^{n}}\right|} = \frac{1}{\ln n}

What is left is a familiar limit, since $\ln n$ grows without bound.

L = \lim_{n \to \infty}\frac{1}{\ln n} = 0 < 1

So the series converges absolutely. With $L = 0$ the tail eventually shrinks faster than any geometric series you care to name.

Why the ratio test is the wrong tool

The ratio drags in two different logarithms and an exponent that refuses to cancel.

\frac{a_{n+1}}{a_{n}} = \frac{(\ln n)^{n}}{\left(\ln (n+1)\right)^{n+1}}

That limit is still $0$ , but reaching it means handling $\left(\frac{\ln n}{\ln (n+1)}\right)^{n}$ , a base near $1$ raised to a growing power. Picking the test is most of the work on a question like this, and the root test is the one built for $n$ th powers.

The sum has to start at n = 2

At $n = 1$ the term is undefined, because $\ln 1 = 0$ and the denominator would vanish. The index has to open at $2$ .

The opening term is bigger than $1$ : $\ln 2 \approx 0.693$ , so $(\ln 2)^{2} \approx 0.48$ and the term is about $2.08$ . Terms only drop below $1$ from $n = 3$ , where $\ln n$ finally passes $1$ .

Large opening terms are allowed

Convergence describes the tail. Changing or deleting finitely many terms at the front can move the value of a sum but can never change whether it converges.

The mistakes students make

Reading the term correctly is most of the battle.

Reading $(\ln n)^{n}$ as $\ln\left(n^{n}\right)$ . The second one is $n \ln n$ , a vastly smaller quantity, and it would give a divergent series.
Starting the sum at $n = 1$ , where the term does not exist.
Concluding divergence because $\sum \frac{1}{\ln n}$ diverges. The exponent $n$ changes the size of the term completely.

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/(ln n)^n converge?

Yes, absolutely. The root test gives $L = 0 < 1$ .

Why does this series start at n = 2?

Because $\ln 1 = 0$ , so the term at $n = 1$ would require division by zero.

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

When the term is an $n$ th power. The root test strips the exponent, while the ratio test leaves you comparing two different bases.