AP Calculus BC

Does the Sum of 1/n^4 Converge? Yes

Q: Does the sum of 1/n^4 converge?

Yes. It is a p-series with p = 4, and every p-series with p > 1 converges.

Q: What does the sum of 1/n^4 equal?

\frac{\pi^{4}}{90}, about 1.0823. That value is a known result, not something the convergence test hands you.

Q: Why does 1/n^4 converge when 1/n does not?

Only the exponent decides it. p = 4 is above the cutoff, p = 1 sits exactly on it, and the boundary case diverges.

The sum of 1 over n to the fourth converges. It is a p-series with p equal to 4, and p above 1 means convergence. The exact value is pi to the fourth over 90, about 1.0823, but that number comes from Euler and not from the p-series test, which only ever returns a verdict.

\sum_{n=1}^{\infty}\frac{1}{n^{4}}

Converges

\text{sum} = \frac{\pi^{4}}{90}

Settled by the p-series test.

Match it to the p-series template

The term is a single power of $n$ in the denominator and nothing else, so the p-series rule applies with no rewriting.

\sum_{n=1}^{\infty}\frac{1}{n^{4}}, \qquad p = 4

A p-series converges when $p > 1$ and diverges when $p \le 1$ . Here $p = 4$ sits well above the line, so the series converges, and the positive terms make that absolute convergence.

The value is Euler's, not the test's

Euler settled the $p = 2$ case, the Basel problem, and the same machinery handles the fourth power.

\sum_{n=1}^{\infty}\frac{1}{n^{4}} = \frac{\pi^{4}}{90} \approx 1.0823

A test returns a verdict, never a value

Nothing inside the p-series test produces pi to the fourth over 90. Piling up terms on a calculator will creep towards 1.0823 and establish nothing, because a partial sum is a finite number and finite numbers cannot rule out a slow divergence. The exponent rule is the entire proof.

The mistakes students make

The errors here are about where $p$ lives and about what a convergence test is for.

Reading $p$ off the numerator or the starting index. In $\frac{1}{n^{4}}$ the value of $p$ is the exponent on $n$ in the denominator, so $p = 4$ .
Quoting $\frac{\pi^{4}}{90}$ as though the p-series test had produced it. The test gives convergence and stops there; the value is a separate and far harder result.
Treating the cutoff as $p \ge 1$ . Convergence needs $p$ strictly greater than $1$ , since $p = 1$ is the harmonic series and it diverges.

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^4 converge?

Yes. It is a p-series with $p = 4$ , and every p-series with $p > 1$ converges.

What does the sum of 1/n^4 equal?

$\frac{\pi^{4}}{90}$ , about $1.0823$ . That value is a known result, not something the convergence test hands you.

Why does 1/n^4 converge when 1/n does not?

Only the exponent decides it. $p = 4$ is above the cutoff, $p = 1$ sits exactly on it, and the boundary case diverges.