AP Calculus BC

Does the Sum of n/3^n Converge? Yes, to 3/4

The sum of n over 3 to the n converges. The ratio test gives a limit of one third. Its exact sum is three quarters, which comes from differentiating the geometric series rather than from the convergence test.

∑n=1∞n3n\sum_{n=1}^{\infty}\frac{n}{3^{n}}

Converges

sum=34\text{sum} = \frac{3}{4}

Settled by the ratio test.

The ratio test

an+1an=n+13n+1⋅3nn=13⋅n+1n⟶13<1\frac{a_{n+1}}{a_{n}} = \frac{n+1}{3^{n+1}}\cdot\frac{3^{n}}{n} = \frac{1}{3}\cdot\frac{n+1}{n} \longrightarrow \frac{1}{3} < 1

The linear numerator is no match for an exponential denominator, and the ratio limit is just the reciprocal of the base.

Where 3/4 comes from

∑n=1∞nxn=x(1−x)2,∣x∣<1\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^{2}}, \qquad |x| < 1

At x=13x = \frac{1}{3} this gives 1/3(2/3)2=1/34/9=34\frac{1/3}{(2/3)^{2}} = \frac{1/3}{4/9} = \frac{3}{4}.

It is not geometric

The factor of n means consecutive terms do not share a constant ratio, only a limiting one. The geometric sum formula does not apply, which is why the value needs the differentiated identity instead.

The mistakes students make

  • Applying the geometric sum formula directly. The extra nn makes the ratio non-constant.
  • Reporting 13\frac{1}{3} as the sum. That is LL, not the total.

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/3^n converge?

Yes. The ratio test gives L=13<1L = \frac{1}{3} < 1.

What is the exact sum?

34\frac{3}{4}, from ∑nxn=x(1−x)2\sum n x^{n} = \frac{x}{(1-x)^{2}} at x=13x = \frac{1}{3}.