AP Calculus BC

Does the Sum of 2^n/n^2 Converge? No

The sum of 2 to the n over n squared diverges. The ratio test gives a limit of 2, which is greater than 1, so the series diverges. The exponential in the numerator grows far faster than the square in the denominator can shrink it.

∑n=1∞2nn2\sum_{n=1}^{\infty}\frac{2^{n}}{n^{2}}

Diverges

Settled by the ratio test.

The ratio test

Exponentials are the signal to reach for the ratio test, because the power cancels cleanly.

an+1an=2n+1(n+1)2⋅n22n=2(nn+1)2\frac{a_{n+1}}{a_{n}} = \frac{2^{n+1}}{(n+1)^{2}}\cdot\frac{n^{2}}{2^{n}} = 2\left(\frac{n}{n+1}\right)^{2}
L=lim⁡n→∞2(nn+1)2=2⋅1=2>1L = \lim_{n \to \infty} 2\left(\frac{n}{n+1}\right)^{2} = 2 \cdot 1 = 2 > 1

A ratio limit above 11 proves divergence. The terms are eventually multiplied by roughly 22 each step, so they grow rather than shrink.

The growth ordering behind it

The nth term test also settles this one, since 2nn2→∞\frac{2^{n}}{n^{2}} \to \infty rather than 00. Exponential growth dominates polynomial growth, which is the ordering BC expects you to know.

Two valid routes

Either the ratio test with L = 2, or the nth term test on terms that grow without bound. When more than one test works, the shorter justification is the one to write down.

The mistakes students make

  • Assuming the n2n^{2} underneath forces convergence. No polynomial can hold back an exponential.
  • Inverting the ratio and getting 12\frac{1}{2}, then concluding convergence. The ratio is an+1an\frac{a_{n+1}}{a_{n}}, newer over older.

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 2^n/n^2 converge?

No. The ratio test gives L=2>1L = 2 > 1, so it diverges.

Could I use the nth term test instead?

Yes. The terms tend to infinity, not 00, which proves divergence in one line.