Telescoping Sums Advanced. This lesson goes beyond core Algebra I. You can skip it.

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.

Look at the third term by itself, $\frac{1}{3\cdot 4}$. It splits into a difference of two simpler fractions,

which you can check by combining the right side over a common denominator: $\frac13 - \frac14 = \frac{4-3}{12} = \frac{1}{12} = \frac{1}{3\cdot 4}$. Every term has this same shape, $\frac{1}{k(k+1)}$, and the identical split works for any $k$, not just $3$:

You do not have to take that on faith either. 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 values of a simpler sequence, one step apart, and when you chain those differences together, the insides cancel in pairs. The same cancellation still works when the values are farther than one step apart, as you’ll see later in this lesson.

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 a few survivors. In general, the method is to find a sequence $b_k$ so that each term of your sum is a shifted difference of $b_k$‘s values, then read the answer straight off as whatever boundary values fail to cancel; a wider gap between the shifted positions leaves more of them to keep and combine.

The simplest case, a gap of $1$, needs only 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 $b_{n+1} - b_1$), since those are the only two boundary values a gap-$1$ telescope leaves. 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 shifted values of some sequence will collapse the same way. Take the sum of $\sqrt{k+1} - \sqrt{k}$, which is already a consecutive 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 down to a small, fixed number of boundary terms: two for a gap-1 telescope like $b_1 - b_{n+1}$, or four for the gap-2 telescope above. Either way, whether the infinite version has a finite value depends entirely on whether the complete surviving boundary expression, not necessarily each piece of it on its own, settles at one fixed value as $n$ grows. Two of our gap-1 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 surviving expression settles at $1$, and the infinite sum is that 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 shape, opposite fate: a telescoping sum has a finite infinite value exactly when its complete surviving boundary expression settles at one fixed value instead of running off. A wider-gap telescope can settle even where an individual survivor does not, as long as the survivors are combined first; what has to settle is the combination, not each piece checked on its own.

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