Telescoping Sums

Sums of fractions like $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots$ do not look like they telescope at first glance, but a partial-fraction-style split turns each term into a difference of two simpler fractions, revealing the same cancellation pattern from the previous lesson. This lesson builds that splitting technique and extends it to sums where the fractions skip over an index.

What you will learn

• Apply partial-fraction-style splitting to a telescoping sum
• Evaluate a finite telescoping sum
• Extend the technique to sums with a shifted index

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

Splitting a fraction for telescoping

A fraction of the form $\frac{1}{k(k+1)}$ splits into a difference of two simpler fractions: $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$. This can be checked directly by combining the right side over a common denominator: $\frac{1}{k}-\frac{1}{k+1}=\frac{(k+1)-k}{k(k+1)}=\frac{1}{k(k+1)}$.

More generally, a fraction $\frac{1}{k(k+d)}$, where the two factors in the denominator differ by $d$, splits as $\frac{1}{k(k+d)}=\frac{1}{d}\left(\frac{1}{k}-\frac{1}{k+d}\right)$. The factor of $\frac{1}{d}$ is required to make the split exact: combining the right side over a common denominator gives $\frac{1}{d}\cdot\frac{d}{k(k+d)}=\frac{1}{k(k+d)}$, confirming the identity.

This split is worth memorizing as a pattern rather than re-deriving every time: look at the two factors multiplied together in the denominator, subtract the smaller from the larger to find $d$, and write the difference of reciprocals scaled by $\frac{1}{d}$ out front. Spotting a denominator that factors into two terms differing by a constant is the entire skill behind recognizing that a fraction sum is secretly telescoping.

Evaluating a finite telescoping sum

Once every term is rewritten as $\frac{1}{k}-\frac{1}{k+1}$, the sum $\sum_{k=1}^{n}\frac{1}{k(k+1)}$ becomes $\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)$, which telescopes exactly as in the previous lesson: every middle fraction cancels, leaving $\frac{1}{1}-\frac{1}{n+1}=1-\frac{1}{n+1}=\frac{n}{n+1}$.

The same process handles a sum that does not start at $k=1$: for $\sum_{k=a}^{b}\frac{1}{k(k+1)}$, the surviving pieces are the first fraction, $\frac{1}{a}$, and the negative of the last fraction, $\frac{1}{b+1}$, giving $\frac{1}{a}-\frac{1}{b+1}$.

Shifted-index sums: when more than one term survives

When the denominator's two factors differ by more than 1, such as $\frac{1}{k(k+2)}$, the split $\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)$ no longer cancels neighbor-to-neighbor, because a term's positive piece $\frac{1}{k}$ only cancels with the negative piece $-\frac{1}{k+2}$ appearing two steps later, not the very next term.

This means two consecutive fractions survive at the start of the sum and two survive at the end, instead of just one on each side. Writing out $\sum_{k=1}^{6}\frac{1}{k(k+2)}=\frac{1}{2}\Big[\left(\tfrac{1}{1}-\tfrac{1}{3}\right)+\left(\tfrac{1}{2}-\tfrac{1}{4}\right)+\left(\tfrac{1}{3}-\tfrac{1}{5}\right)+\left(\tfrac{1}{4}-\tfrac{1}{6}\right)+\left(\tfrac{1}{5}-\tfrac{1}{7}\right)+\left(\tfrac{1}{6}-\tfrac{1}{8}\right)\Big]$ shows $\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}$ each cancelling against a matching negative piece two terms later, leaving only $\frac{1}{1}+\frac{1}{2}-\frac{1}{7}-\frac{1}{8}$ inside the brackets.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Splitting $\frac{1}{k(k+d)}$ without the required factor of $\frac{1}{d}$ in front.
• Assuming only one term survives at each end of a shifted-index sum, when a gap of $d>1$ actually leaves $d$ terms surviving on each side.
• Using the wrong last index when the sum starts above $k=1$, forgetting that the surviving negative piece is $\frac{1}{b+1}$ for a sum ending at $k=b$.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• A fraction $\frac{1}{k(k+d)}$ splits as $\frac{1}{d}\left(\frac{1}{k}-\frac{1}{k+d}\right)$; for $d=1$ this is simply $\frac{1}{k}-\frac{1}{k+1}$.
• After splitting, $\sum_{k=a}^{b}\frac{1}{k(k+1)}$ telescopes to $\frac{1}{a}-\frac{1}{b+1}$.
• When the denominator gap $d$ is greater than 1, exactly $d$ fractions survive at the start and $d$ at the end, not just one on each side.