AP Calculus BC

Does the Sum of 1/e^n Converge? Yes

The sum of 1 over e to the n converges absolutely. Rewriting the term as 1 over e all raised to the n shows it is geometric with common ratio about 0.368, comfortably under 1. The sum collapses to 1 over e minus 1, roughly 0.582.

∑n=1∞1en\sum_{n=1}^{\infty}\frac{1}{e^{n}}

Converges

sum=1e−1\text{sum} = \frac{1}{e-1}

Settled by the geometric series test.

A power of e is still geometric

The base is fixed and the exponent carries the nn, which is the definition of geometric. An irrational base is no obstacle.

1en=(1e)n,r=1e≈0.3679\frac{1}{e^{n}} = \left(\frac{1}{e}\right)^{n}, \qquad r = \frac{1}{e} \approx 0.3679

Because ∣r∣<1|r| < 1 the series converges, and it converges fast: each term is about 2.72.7 times smaller than the one before.

The sum tidies up to 1/(e-1)

The first term is 1e\frac{1}{e} and the ratio is 1e\frac{1}{e}. Multiplying the top and bottom of a1−r\frac{a}{1-r} by ee clears the nested fractions.

∑n=1∞(1e)n=1e1−1e=1e−1≈0.5820\sum_{n=1}^{\infty}\left(\frac{1}{e}\right)^{n} = \frac{\frac{1}{e}}{1-\frac{1}{e}} = \frac{1}{e-1} \approx 0.5820

1/e^n and 1/n^e are different animals

In 1 over e to the n the variable sits in the exponent, so the series is geometric. In 1 over n to the e the variable sits in the base, so it is a p-series with p of about 2.718. Both converge, by different tests, and only the geometric one hands you a sum.

The mistakes students make

Recognition is the whole difficulty on this one.

  • Not seeing e−ne^{-n} as geometric and going straight to the integral test. That test does work, but the geometric form gives the exact sum as well as the verdict.
  • Using a=1a = 1, which is the n=0n = 0 term. Starting at n=1n = 1 gives a=1ea = \frac{1}{e} and a sum of 1e−1\frac{1}{e-1}, while starting at n=0n = 0 would give ee−1\frac{e}{e-1}.
  • Reading 1en\frac{1}{e^{n}} as 1ne\frac{1}{n^{e}}. The first is geometric, the second is a p-series.

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 1/e^n converge?

Yes. It is geometric with r=1e≈0.368r = \frac{1}{e} \approx 0.368, and ∣r∣<1|r| < 1 gives convergence.

What does the sum of 1/e^n equal?

Starting at n=1n = 1, the sum is 1e−1\frac{1}{e-1}, about 0.5820.582.

Is 1/e^n geometric or a p-series?

Geometric. The nn is in the exponent, so successive terms share a fixed ratio of 1e\frac{1}{e}.