AP Calculus BC

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

The sum of 1 over n times n plus 1 converges to exactly 1. Partial fractions splits the term into 1 over n minus 1 over n plus 1, so consecutive pieces cancel and the partial sum collapses to 1 minus 1 over N plus 1.

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

Converges

sum=1\text{sum} = 1

Settled by telescoping partial sums.

Split, then watch it collapse

Partial fractions is what exposes the structure.

1n(n+1)=1n−1n+1\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}

Writing out the partial sum, every interior piece cancels against its neighbour.

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

Taking the limit gives the sum exactly.

lim⁡N→∞(1−1N+1)=1\lim_{N \to \infty}\left(1 - \frac{1}{N+1}\right) = 1

One of the few series with an easy exact sum

Telescoping and geometric are the two families where the AP exam expects a value rather than a verdict. Everything else is a yes or no question.

Work with the finite sum first

Cancelling directly inside an infinite sum is not justified. Write the partial sum, cancel there, then take the limit. That order is what makes the argument legitimate.

The mistakes students make

  • Cancelling inside the infinite sum without forming SNS_{N} first.
  • Losing the surviving endpoints. After cancellation the first and last pieces remain, and they are what produce the answer.
  • Assuming every series with a nice term telescopes. It requires the partial fraction split to leave consecutive, opposite pieces.

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+1)) converge?

Yes, to exactly 11.

Why does it telescope?

Because partial fractions turns each term into 1n−1n+1\frac{1}{n} - \frac{1}{n+1}, and consecutive terms cancel.

Which series have easy exact sums?

Geometric and telescoping. For almost everything else a test gives only convergence or divergence.