AP Calculus BC

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

The sum of n squared over 2 to the n converges absolutely. Dividing one term by the one before it leaves one half times the square of n plus 1 over n. That square settles at 1, so the ratio limit is one half, comfortably under the cutoff of 1.

∑n=1∞n22n\sum_{n=1}^{\infty}\frac{n^{2}}{2^{n}}

Converges

Settled by the ratio test.

Set up the ratio

A polynomial above the line and a constant raised to the power n below it is the clearest signal there is for the ratio test.

an+1an=(n+1)22n+1⋅2nn2=12(n+1n)2\frac{a_{n+1}}{a_{n}} = \frac{(n+1)^{2}}{2^{n+1}}\cdot\frac{2^{n}}{n^{2}} = \frac{1}{2}\left(\frac{n+1}{n}\right)^{2}

The bracket closes in on 1, so the constant factor is all that survives.

L=lim⁡n→∞12(1+1n)2=12L = \lim_{n \to \infty}\frac{1}{2}\left(1 + \frac{1}{n}\right)^{2} = \frac{1}{2}

With L=12<1L = \frac{1}{2} < 1 the ratio test gives absolute convergence, and that limit is the entire justification.

The polynomial only delays the collapse

Each step doubles the denominator while multiplying the numerator by (n+1n)2\left(\frac{n+1}{n}\right)^{2}, a factor that drops towards 11. Doubling wins, and it wins against every fixed power of nn.

The early terms actually grow: 12\frac{1}{2}, then 11, then 98\frac{9}{8}, before they turn over and fall. A larger power of nn pushes that turning point further out and changes nothing else.

The tail is what matters

Convergence is a statement about the far end of the series. Terms that rise for a while, or a partial sum that looks large, have no bearing on the verdict. The ratio limit does.

The mistakes students make

The ratio test is mechanical, and the slips are mechanical too.

  • Forgetting to invert ana_{n}. The 2n2^{n} moves upstairs, giving 2n2n+1=12\frac{2^{n}}{2^{n+1}} = \frac{1}{2}, not 22.
  • Taking the first few rising terms as evidence of divergence.
  • Checking an+1an<1\frac{a_{n+1}}{a_{n}} < 1 at one value of nn and stopping. The test is a statement about the limit.

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^2/2^n converge?

Yes, absolutely. The ratio test gives L=12<1L = \frac{1}{2} < 1.

Why is the ratio test the right choice here?

Because the term contains 2n2^{n}. Powers of a constant cancel cleanly in an+1an\frac{a_{n+1}}{a_{n}} and leave a limit you can read off.

Does the power of n change the answer?

No. n2n\frac{n}{2^{n}}, n22n\frac{n^{2}}{2^{n}} and n102n\frac{n^{10}}{2^{n}} all give L=12L = \frac{1}{2}.