AP Calculus BC

Does the Sum of n^4/2^n Converge? Yes

The series converges. The ratio test gives a limit of 1/2, which is less than one. The general lesson is that an exponential in the denominator beats any fixed polynomial power in the numerator, no matter how large that power is.

∑n=1∞n42n\sum_{n=1}^{\infty}\frac{n^{4}}{2^{n}}

Converges

Settled by the ratio test.

Working the ratio

∣an+1an∣=(n+1)42n+1⋅2nn4=12(n+1n)4\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^{4}}{2^{n+1}} \cdot \frac{2^{n}}{n^{4}} = \frac{1}{2}\left(\frac{n+1}{n}\right)^{4}

The bracket tends to 1, so the whole thing tends to 1/21/2. Since 1/2<11/2 < 1, the ratio test proves convergence, and it proves absolute convergence at that.

Why the power does not matter

Replace the 4 by 40 and the ratio becomes 12(n+1n)40\frac{1}{2}\left(\frac{n+1}{n}\right)^{40}, which still tends to 1/21/2. The polynomial contributes a factor that goes to 1 whatever its degree, so the exponential base alone decides.

The terms do grow at first: n4/2nn^{4}/2^{n} climbs until about n=6n = 6 before turning over. Early growth is not evidence of divergence, because convergence is decided by the tail.

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

Why does the ratio test suit this series?

The terms mix a polynomial with an exponential, and the ratio cancels both cleanly: the exponential leaves a constant and the polynomial leaves a factor tending to 1.

Would the p-series test work here?

No. This is not a pure power of n, so the p-series test does not apply. The presence of 2n2^{n} is what makes the ratio test the right tool.