AP Calculus BC

Does the Sum of 1/(n(n+2)) Converge? Yes, to 3/4

The sum of 1 over n times n plus 2 converges to exactly 3/4. Partial fractions gives one half times 1 over n minus 1 over n plus 2. The gap of two means two front terms survive, 1 and 1 over 2, so the total is one half of 3/2.

∑n=1∞1n(n+2)\sum_{n=1}^{\infty}\frac{1}{n(n+2)}

Converges

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

Settled by telescoping partial sums.

Partial fractions, with a factor of one half

The denominator is already factored, so the split is quick. Because the two factors sit two apart rather than one apart, a 12\frac{1}{2} comes out in front.

1n(n+2)=12(1n−1n+2)\frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)

Each negative piece −1n+2-\frac{1}{n+2} is cancelled by a positive piece two steps later, not one step later. That single difference changes what is left over at the ends.

Two terms survive at the front

Write the finite sum and track what fails to find a partner.

SN=12(1+12−1N+1−1N+2)S_{N} = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right)

Nothing cancels the 11 or the 12\frac{1}{2}, because the pieces that would do so sit before the series begins. At the other end, two pieces wait for partners that never arrive.

Both trailing pieces tend to 00, so the limit is decided by the front pair.

lim⁡N→∞SN=12(1+12)=34\lim_{N \to \infty} S_{N} = \frac{1}{2}\left(1 + \frac{1}{2}\right) = \frac{3}{4}

How many terms survive, in general

For ∑1n(n+k)\sum \frac{1}{n(n+k)} the split leaves a gap of kk, and exactly kk terms survive at each end. The constant out front is 1k\frac{1}{k}, not 12\frac{1}{2}, and the survivors are the first kk reciprocals.

1n(n+k)=1k(1n−1n+k),∑n=1∞1n(n+k)=1k(1+12+⋯+1k)\frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right), \qquad \sum_{n=1}^{\infty}\frac{1}{n(n+k)} = \frac{1}{k}\left(1 + \frac{1}{2} + \cdots + \frac{1}{k}\right)

With k=1k = 1 that gives 11. With k=2k = 2 it gives 12(1+12)=34\frac{1}{2}\left(1 + \frac{1}{2}\right) = \frac{3}{4}, the case on this page. With k=3k = 3 it gives 13(1+12+13)=1118\frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{3}\right) = \frac{11}{18}, so do not carry the 12\frac{1}{2} from this page into another gap.

Cancel inside the finite sum

Form the partial sum, cancel there, then take the limit. That order is what makes the argument legal. Adding terms on a calculator can hint at a value, but it never establishes one.

The mistakes students make

Almost every error on this series happens in the first two lines of work.

  • Dropping the 12\frac{1}{2} from the split. Solving 1=A(n+2)+Bn1 = A(n+2) + Bn gives A=12A = \frac{1}{2} and B=−12B = -\frac{1}{2}, not 11 and −1-1.
  • Keeping only one surviving front term, out of habit from ∑1n(n+1)\sum \frac{1}{n(n+1)}. A gap of two leaves both 11 and 12\frac{1}{2}.
  • Omitting the two trailing pieces from SNS_{N}. They do vanish in the limit, but the formula for SNS_{N} is wrong without them.

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/(n(n+2)) converge?

Yes, to exactly 34\frac{3}{4}, and telescoping gives both the verdict and the value.

Why is the sum 3/4 and not 1?

The gap of two leaves two front terms, 11 and 12\frac{1}{2}, and the partial fraction split carries a factor of 12\frac{1}{2}. So the total is 12⋅32=34\frac{1}{2}\cdot\frac{3}{2} = \frac{3}{4}.

How do I know a series telescopes?

Split the term with partial fractions. If the pieces are the same expression at two shifted indices, consecutive terms cancel and only the ends remain.