AP Calculus BC

Does the Sum of 2^(n+1)/3^n Converge? Yes

The sum of 2 to the n plus 1 over 3 to the n converges absolutely. The extra 1 in the exponent is a constant factor of 2 in disguise, so the common ratio is still 2 over 3. Pulling that 2 out front leaves a first term of 2 over 3, the geometric sum that remains is 2, and 2 times that is exactly 4.

∑n=1∞2n+13n\sum_{n=1}^{\infty}\frac{2^{n+1}}{3^{n}}

Converges

sum=4\text{sum} = 4

Settled by the geometric series test.

Peel the constant off first

Split the shifted exponent before classifying anything: 2n+12^{n+1} is just 22 times 2n2^{n}.

2n+13n=2⋅2n3n=2(23)n\frac{2^{n+1}}{3^{n}} = 2\cdot\frac{2^{n}}{3^{n}} = 2\left(\frac{2}{3}\right)^{n}

A constant multiple cannot change whether a series converges, so this is geometric with r=23r = \frac{2}{3}, and ∣r∣<1|r| < 1 gives convergence.

Two routes to the same 4

Pull the constant outside the sum and apply the formula to what remains.

∑n=1∞2n+13n=2∑n=1∞(23)n=2⋅231−23=2⋅2=4\sum_{n=1}^{\infty}\frac{2^{n+1}}{3^{n}} = 2\sum_{n=1}^{\infty}\left(\frac{2}{3}\right)^{n} = 2\cdot\frac{\frac{2}{3}}{1-\frac{2}{3}} = 2\cdot 2 = 4

Or leave the 22 inside. The first term at n=1n = 1 is 43\frac{4}{3}, and 4/31−2/3=4\frac{4/3}{1-2/3} = 4 again. Agreement between the two routes is a good check that you read the first term correctly.

The mistakes students make

A shifted exponent creates bookkeeping errors, not a new test.

  • Believing the +1+1 changes the ratio. Divide consecutive terms: 2n+23n+1⋅3n2n+1=23\frac{2^{n+2}}{3^{n+1}}\cdot\frac{3^{n}}{2^{n+1}} = \frac{2}{3} for every nn.
  • Factoring the 22 out and then forgetting to multiply it back in, which reports 22 instead of 44.
  • Using a=1a = 1 out of habit and reporting 33. The first term of this series is 43\frac{4}{3}.

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+1)/3^n converge?

Yes, to 44. It is geometric with r=23r = \frac{2}{3}.

How do you deal with the n+1 in the exponent?

Write 2n+1=2⋅2n2^{n+1} = 2 \cdot 2^{n} and move the 22 outside the sum. What is left is a plain geometric series.

What is the first term of this series?

At n=1n = 1 the term is 2231=43\frac{2^{2}}{3^{1}} = \frac{4}{3}, which is the aa in a1−r\frac{a}{1-r}.