AP Calculus BC

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

The sum of n squared over n factorial converges. The ratio test gives a limit of 0, because the factorial in the denominator grows faster than any polynomial in the numerator. No power of n can compete with n factorial.

∑n=1∞n2n!\sum_{n=1}^{\infty}\frac{n^{2}}{n!}

Converges

Settled by the ratio test.

Polynomial over factorial

an+1an=(n+1)2(n+1)!⋅n!n2=(n+1)2n2(n+1)=n+1n2\frac{a_{n+1}}{a_{n}} = \frac{(n+1)^{2}}{(n+1)!}\cdot\frac{n!}{n^{2}} = \frac{(n+1)^{2}}{n^{2}(n+1)} = \frac{n+1}{n^{2}}
L=lim⁡n→∞n+1n2=0<1L = \lim_{n \to \infty}\frac{n+1}{n^{2}} = 0 < 1

The same argument works for nkn^{k} for any fixed kk: the ratio ends up as a polynomial over a polynomial of higher degree, which tends to 00.

The growth ordering

ln⁡n  ≪  nk  ≪  an  ≪  n!  ≪  nn\ln n \;\ll\; n^{k} \;\ll\; a^{n} \;\ll\; n! \;\ll\; n^{n}

Each item is beaten by the next. Knowing this ordering answers most convergence questions before any test is run, and it is why a factorial denominator is nearly always good news.

The mistakes students make

  • Cancelling (n+1)2n2\frac{(n+1)^{2}}{n^{2}} to 11 too early and losing the 1n+1\frac{1}{n+1} that actually drives the limit.
  • Assuming the squared numerator changes the verdict. It does not; no power does.

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 n^2/n! converge?

Yes, by the ratio test with L=0L = 0.

What about n^10/n!?

Also converges. The ratio still tends to 00 for any fixed power.