AP Calculus BC

Does the Sum of 1/(n(n+3)) Converge? Yes, to 11/18

The series converges to exactly 11/18. Partial fractions turn it into a telescoping sum, and because the gap between the factors is three rather than one, three terms survive the cancellation instead of just one.

∑n=1∞1n(n+3)\sum_{n=1}^{\infty}\frac{1}{n(n+3)}

Converges

sum=1118\text{sum} = \frac{11}{18}

Settled by telescoping partial sums.

Splitting the term

1n(n+3)=13(1n−1n+3)\frac{1}{n(n+3)} = \frac{1}{3}\left(\frac{1}{n} - \frac{1}{n+3}\right)

The factor of one third is easy to lose. It appears because the two factors differ by 3: in general 1n(n+k)=1k(1n−1n+k)\frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right).

Verify by recombining over a common denominator: 1n−1n+3=3n(n+3)\frac{1}{n} - \frac{1}{n+3} = \frac{3}{n(n+3)}, and dividing by 3 returns the original term.

What survives the cancellation

Writing out the partial sum, each −1n+3-\frac{1}{n+3} cancels against the 1n\frac{1}{n} three places later. Three terms at the front never get cancelled, and three at the back have nothing to cancel against yet.

SN=13(1+12+13−1N+1−1N+2−1N+3)S_N = \frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{N+1} - \frac{1}{N+2} - \frac{1}{N+3}\right)

The three trailing pieces vanish as NN grows, leaving 13(1+12+13)=13⋅116=1118\frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{3}\right) = \frac{1}{3}\cdot\frac{11}{6} = \frac{11}{18}.

With a gap of 1, as in ∑1n(n+1)\sum \frac{1}{n(n+1)}, only one term survives and the sum is 1. The size of the gap sets how many leftovers there are, so writing out the first few terms explicitly is worth the time.

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

How many terms survive a telescoping sum?

As many as the gap between the factors. A gap of k leaves k terms at the front, and k at the back that vanish in the limit.

Why does the one third appear?

Partial fractions over factors differing by k always produce a factor of 1/k1/k. Recombining the two fractions gives k over the original denominator, so dividing by k restores the term.