Telescoping Sums

A sum that seems to have no total

Take the fractions $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \cdots$ and just add them one at a time, keeping a running total after each new term:

The running totals are $\frac12, \frac23, \frac34, \frac45$, and the pattern is impossible to miss: after $n$ terms the sum appears to be $\frac{n}{n+1}$. That is a bold guess from four data points, and guessing is not proving. The reason it is true comes from rewriting a single term.

Every term has the shape $\frac{1}{k(k+1)}$, and this fraction can be split into a difference of two simpler fractions:

You do not have to take that on faith. Combine the right side back over the common denominator $k(k+1)$ and watch it rebuild the left side exactly:

The two sides agree, so the split is an exact identity, not an approximation. Recombining like this is how you check any split in this lesson: if the pieces add back to the original, the split is safe to use.

Now replace each term of the sum with its split and see what happens:

The $-\frac12$ in the first bracket meets the $+\frac12$ in the second and cancels. The $-\frac13$ meets the $+\frac13$ in the bracket after it. Every fraction on the inside is subtracted once and added back once, so it disappears. Only two pieces have no partner: the $\frac11$ at the very front and the $-\frac{1}{n+1}$ at the very end:

That is the guessed pattern, now proved. The messy sum of $n$ fractions is really just $1 - \frac{1}{n+1}$ in disguise.

The telescoping principle

The workhorse example used one idea, and that idea works far beyond this single sum. Each term was a difference of two consecutive values of a simpler sequence, and when you chain those differences together, the insides cancel in pairs.

This is where the name comes from. An old pocket telescope is a stack of tubes that slides shut into a short ring. A telescoping sum does the same thing: a line of many terms slides shut into just two. The entire method is two steps. First, find a sequence $b_k$ so that each term of your sum is $b_k - b_{k+1}$ (or $b_{k+1} - b_k$). Second, read the answer straight off as $b_1 - b_{n+1}$ (or $b_{n+1} - b_1$). For the workhorse sum, $b_k = \frac1k$, so $b_1 - b_{n+1} = 1 - \frac{1}{n+1}$, exactly what we found by hand.

When two terms survive at each end

The gap between the two factors is not always $1$. Consider $\frac{1}{k(k+2)}$, where the factors differ by $2$. The natural split is again a difference, but a plain $\frac1k - \frac{1}{k+2}$ overshoots, so a correcting factor is needed. Combine $\frac1k - \frac{1}{k+2}$ and see why:

That is twice too big, so halve it. The correct split carries a factor of $\frac12$:

Because the gap is $2$, each negative piece cancels the positive piece two brackets later, not one. That leaves a longer tail of survivors at both ends, which is the classic trap in these problems.

Telescoping with square roots

Telescopes are not only for fractions. Any expression that is a difference of consecutive values will collapse the same way. Take the sum of $\sqrt{k+1} - \sqrt{k}$, which is already a difference with $b_k = \sqrt{k}$ built in the mirror direction $b_{k+1} - b_k$:

The interior square roots all cancel, and the mirror form leaves the last value positive and the first negative, giving $\sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1$.

You will also meet this same telescope disguised as a fraction. The term $\frac{1}{\sqrt{k} + \sqrt{k+1}}$ looks nothing like a difference, but rationalizing the denominator (the trick from the radicals chapter, multiply top and bottom by the conjugate $\sqrt{k+1} - \sqrt{k}$) uncovers one:

The denominator became a difference of squares, $(k+1) - k = 1$, so the whole fraction is just $\sqrt{k+1} - \sqrt{k}$, the very telescope above.

Differences of squares

A telescope can also be hiding inside a sum of whole numbers. The difference of two consecutive squares is

so the odd number $2k+1$ is secretly the difference $b_{k+1} - b_k$ with $b_k = k^{2}$. Summing the odd numbers therefore telescopes:

Reading the terms out, $2k+1$ runs through $3, 5, 7, \ldots, 2n+1$, so this says $3 + 5 + 7 + \cdots + (2n+1) = (n+1)^{2} - 1$. Add the missing $1$ to both sides and you recover the familiar fact that the first $n+1$ odd numbers sum to a perfect square, $(n+1)^{2}$. The old “odd numbers build squares” pattern is a telescope in disguise.

The same difference idea reaches beyond fractions: the log-quotient rule writes $\log\!\left(\frac{k+1}{k}\right) = \log(k+1) - \log(k)$, so $\sum_{k=1}^{n} \log\!\left(\frac{k+1}{k}\right)$ collapses to $\log(n+1)$.

Does the infinite sum settle?

Every telescope in this lesson collapses to $b_1 - b_{n+1}$ or its mirror, so whether the infinite version has a finite value depends entirely on the surviving tail term. Two of our examples show the two outcomes side by side. For $\frac{1}{k(k+1)}$ the finite total is $1 - \frac{1}{n+1}$, and the tail $\frac{1}{n+1}$ shrinks toward nothing, so the infinite sum settles at a clean $1$. For $\sqrt{k+1} - \sqrt{k}$ the finite total is $\sqrt{n+1} - 1$, and $\sqrt{n+1}$ grows without bound as $n$ climbs, so that infinite sum has no finite value at all. Same collapse, opposite fate: a telescoping sum has a finite infinite value exactly when its surviving tail term settles down instead of running off.

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