Telescoping Series

Split each term and the neighbors erase each other in this scene. Only the very first piece and a vanishing tail survive.

Each term breaks into a difference of two fractions. The back half of one term matches the front half of the next, so they cancel. After the erasing, only the opening 1 remains. The tail shrinks away to nothing. So the total is exactly 1.

For example, climb one step of a 🪜 staircase and step back down: you are where you started. Chain these ups and downs together and only the starting and ending positions count.

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

Examples & Exercises

Prove ∑n=1N1n(n+1)=NN+1\sum_{n=1}^{N}\frac{1}{n(n+1)} = \frac{N}{N+1}, so the series sums to 11.

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}
N=24N = 24: S24=24/25=0.96S_{24} = 24/25 = 0.96, only the tail 1/251/25 remains.
Show proof

  1. 1n(n+1)=1n−1n+1\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1} (check by clearing denominators).

  2. Summing telescopes the neighbors away: SN=1−1N+1=NN+1S_N = 1 - \frac{1}{N+1} = \frac{N}{N+1}.

  3. As N→∞N \to \infty, 1N+1→0\frac{1}{N+1} \to 0, so the series sums to 11.

Compute

Prove ∑n=1∞1n(n+2)=34\sum_{n=1}^{\infty}\frac{1}{n(n+2)} = \frac{3}{4}.

1n(n+2)=12(1n−1n+2)\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)
Cancellation skips one neighbor this time.
Show proof

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

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

  3. As N→∞N \to \infty the tail vanishes, leaving 12⋅32=34\frac{1}{2}\cdot\frac{3}{2} = \frac{3}{4}.

Answer

34\frac{3}{4}

Show ∑n=1∞(n+1−n)\sum_{n=1}^{\infty}\left(\sqrt{n+1}-\sqrt{n}\right) diverges.

SN=N+1−1→∞S_N = \sqrt{N+1}-1 \to \infty
Telescoping — but the survivor goes to infinity.
Show proof

  1. SN=(2−1)+(3−2)+⋯+(N+1−N)=N+1−1S_N = (\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\cdots+(\sqrt{N+1}-\sqrt{N}) = \sqrt{N+1}-1.

  2. Since N+1→∞\sqrt{N+1} \to \infty, the partial sums are unbounded.

  3. Telescoping does not always converge: a general telescoping sum SN=bN+1−b1S_N = b_{N+1} - b_1 converges if and only if (bn)(b_n) converges to a finite limit.

    Here bn=n→∞b_n = \sqrt{n} \to \infty, so it diverges.

    (The criterion "the surviving term must tend to 00" is false — the previous problem's SN=1−1N+1→1≠0S_N = 1 - \frac{1}{N+1} \to 1 \ne 0 is a convergent counterexample.)

7. Series