AP Calculus BC

Does the Sum of n/4^n Converge? Yes, to 4/9

The series converges, and its sum is exactly 4/9. The ratio test gives a limit of 1/4, and the exact value comes from differentiating the geometric series formula, a trick that handles any series of the form n times r to the n.

∑n=1∞n4n\sum_{n=1}^{\infty}\frac{n}{4^{n}}

Converges

Settled by the ratio test.

The verdict

∣an+1an∣=n+14n+1⋅4nn=14⋅n+1n⟶14\left|\frac{a_{n+1}}{a_n}\right| = \frac{n+1}{4^{n+1}}\cdot\frac{4^{n}}{n} = \frac{1}{4}\cdot\frac{n+1}{n} \longrightarrow \frac{1}{4}

A limit of 1/4<11/4 < 1 gives absolute convergence. The linear factor nn contributes only n+1n→1\frac{n+1}{n} \to 1 and never changes the answer.

The differentiation trick

Start from the geometric series ∑n=0∞xn=11−x\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x} for ∣x∣<1|x| < 1, and differentiate both sides with respect to xx.

∑n=1∞nxn−1=1(1−x)2⟹∑n=1∞nxn=x(1−x)2\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^{2}} \quad\Longrightarrow\quad \sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^{2}}

Putting x=1/4x = 1/4 gives 1/4(3/4)2=1/49/16=49\frac{1/4}{(3/4)^{2}} = \frac{1/4}{9/16} = \frac{4}{9}. Differentiating a power series term by term is legitimate strictly inside the radius of convergence, and 1/41/4 is well inside.

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

How do I sum a series with n times r to the n?

Differentiate the geometric series. The identity ∑nxn=x/(1−x)2\sum n x^{n} = x/(1-x)^{2} handles every case, and differentiating once more handles n2xnn^{2} x^{n}.

Is term-by-term differentiation always allowed?

Strictly inside the radius of convergence, yes, and the radius is unchanged. At an endpoint it can fail, so endpoints have to be checked separately.