AP Calculus BC

Does the Sum of 1/5^n Converge? Yes, to 1/4

The series converges, and its sum is exactly 1/4. It is geometric with common ratio 1/5, and any geometric series whose ratio is smaller than one in absolute value converges to its first term divided by one minus the ratio.

∑n=1∞15n\sum_{n=1}^{\infty}\frac{1}{5^{n}}

Converges

sum=14\text{sum} = \frac{1}{4}

Settled by the geometric series test.

Reading the ratio off the terms

Divide any term by the one before it: 1/5n+11/5n=15\frac{1/5^{n+1}}{1/5^{n}} = \frac{1}{5}. That quotient is the same number every time, which is what makes the series geometric rather than merely shrinking.

Because ∣r∣=1/5<1|r| = 1/5 < 1, the powers rnr^{n} collapse toward 0 and the partial sums close in on a finite total.

Getting the sum right

The formula is first term over one minus the ratio, and the first term is whatever the series actually starts with. Here the index starts at n=1n = 1, so the first term is 1/51/5, not 1.

1/51−1/5=1/54/5=14\frac{1/5}{1 - 1/5} = \frac{1/5}{4/5} = \frac{1}{4}

Starting the index at n=0n = 0 instead would add a term of 1 and give a sum of 5/45/4. The convergence verdict would not change, because dropping or adding finitely many terms never affects whether a series converges.

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

What is the common ratio of this series?

It is 1/51/5. Divide any term by the previous one and you get 1/51/5 every time, which is exactly what identifies a geometric series.

Why is the sum 1/4 and not 5/4?

Because the index starts at n=1n = 1, so the first term present is 1/51/5. The value 5/45/4 is the sum when the series starts at n=0n = 0 and includes a leading term of 1.