Telescoping Sums: Free Response

Substitute $k=1,2,\ldots,14$ into the given split, one bracket per term.

$\sum_{k=1}^{14}\frac{1}{k(k+1)} = \left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1{14}-\frac1{15}\right).$

Each interior fraction from $\frac12$ through $\frac1{14}$ appears once with a minus sign and once with a plus sign in the very next bracket, so writing the chain out this way is what makes the cancellation visible before any arithmetic is done.

Every interior fraction cancels in pairs, leaving only the front piece $1$ and the back piece $-\frac1{15}$.

$\sum_{k=1}^{14}\frac{1}{k(k+1)} = 1-\frac1{15} = \frac{14}{15}.$

Track a single interior value, say $b_5=\frac15$. It appears as $-\frac15$ in the bracket for $k=4$ and as $+\frac15$ in the bracket for $k=5$, so the two occurrences cancel. The same argument applies to every value strictly between $b_1$ and $b_{n+1}$: each one is subtracted in exactly one bracket and added back in the very next one.

$\sum_{k=1}^{n}(b_k-b_{k+1}) = b_1-b_{n+1}.$

Only the value at the very front, $b_1$, is never subtracted, and only the value at the very back, $-b_{n+1}$, is never added back, so those are the only two pieces that can possibly survive, regardless of how large $n$ is.

Use the common denominator $(k+1)(k+3)$ and subtract the numerators.

$\frac1{k+1}-\frac1{k+3} = \frac{(k+3)-(k+1)}{(k+1)(k+3)} = \frac{2}{(k+1)(k+3)}.$

Part A combined that candidate's right side to $\frac{2}{(k+1)(k+3)}$, which is twice the target fraction $\frac1{(k+1)(k+3)}$, not equal to it. Since the two factors in the denominator differ by $2$, this is exactly the gap-2 pattern: the plain difference always overshoots by a factor of $2$, so halving it fixes the split.

$\frac{1}{(k+1)(k+3)} = \frac12\left(\frac1{k+1}-\frac1{k+3}\right).$

The candidate split, as stated with no $\frac12$, is false.

Combine the proposed right side over the common denominator $k(k+1)$.

$\frac1{k+1}-\frac1k = \frac{k-(k+1)}{k(k+1)} = \frac{-1}{k(k+1)}.$

That is the negative of the target fraction $\frac1{k(k+1)}$, not the fraction itself, so the proposed split is false as written. Nothing is wrong with the pair of fractions chosen, only their order: swapping them, $\frac1k-\frac1{k+1}$, recombines to exactly $\frac1{k(k+1)}$, the correct split. This is the same sign flip that swapping a telescoping sum's direction produces in its final total.

Each term is already a difference of consecutive square roots, $b_{k+1}-b_k$ with $b_k=\sqrt{k}$, so the mirror collapse applies.

$\sum_{k=1}^{n}\left(\sqrt{k+1}-\sqrt{k}\right) = \sqrt{n+1}-\sqrt{1} = \sqrt{n+1}-1.$

As $n$ climbs, $\sqrt{n+1}$ climbs right along with it, with no bound on how large it gets, so the closed form has no limiting value.

The two factors differ by $1$, so this splits cleanly with no correcting factor, $\frac1{(k+2)(k+3)} = \frac1{k+2}-\frac1{k+3}$, exactly the workhorse pattern shifted by $2$. Combining that back confirms it.

$\frac1{k+2}-\frac1{k+3} = \frac{(k+3)-(k+2)}{(k+2)(k+3)} = \frac{1}{(k+2)(k+3)}.$

With $b_k=\frac1{k+2}$, the sum collapses to $b_1-b_{n+1}$.

$\sum_{k=1}^{n}\frac1{(k+2)(k+3)} = \frac13-\frac1{n+3}.$

As $n$ grows, the leftover term $\frac1{n+3}$ shrinks toward zero, so the closed form settles right next to $\frac13$.

A single counterexample is enough to break a claim about EVERY telescoping sum, and part A supplies one: it is an infinite telescoping sum, built exactly the way the student describes, and its closed form $\sqrt{n+1}-1$ has no finite value because the surviving tail $\sqrt{n+1}$ never stops growing.

That alone does not mean NO infinite telescoping sum settles, and part B rules that overcorrection out: its closed form $\frac13-\frac1{n+3}$ approaches a fixed value because its surviving tail $\frac1{n+3}$ shrinks to zero instead. The two tails behave oppositely as $n$ grows.

$\sqrt{n+1}-1 \;\longrightarrow\; \infty, \qquad \frac13-\frac1{n+3} \;\longrightarrow\; \frac13.$

Putting the two together pins down what actually decides the outcome: not whether a sum telescopes (both of these do), but whether the leftover tail term shrinks to zero or not. The claim's honest, guarded form is that an infinite telescoping sum has a finite value exactly when its surviving tail term shrinks to zero.

Test each line against the one directly above it.

Line 1 is the given split for a gap-2 fraction: sound, since the two factors differ by $2$.

Line 2 expands the sum from $k=4$ correctly: sound.

Line 3 claims that only $\frac14$ survives at the front and only $-\frac1{n+2}$ survives at the back. But with a gap of $2$, the negative piece from the $k=4$ bracket, $-\frac16$, does not cancel until the $k=6$ bracket, TWO brackets later, not the very next one:

$\underbrace{-\frac16}_{k=4\text{ bracket}} \quad \text{cancels with} \quad \underbrace{+\frac16}_{k=6\text{ bracket}}.$

That means the negative piece from $k=5$, $-\frac17$, also needs a bracket two steps later to cancel. So the first TWO positive pieces, $\frac14$ and $\frac15$, never find a partner within the sum, and likewise the LAST two negative pieces survive at the back. Line 3 is the first line that does not validly follow.

Line 4 carries out correct algebra, but only on the flawed claim Line 3 handed it, so it inherits the error.

With both front survivors and both back survivors kept, the sum is

$\sum_{k=4}^{n}\frac{1}{k(k+2)} = \frac12\left(\frac14+\frac15-\frac1{n+1}-\frac1{n+2}\right).$

Combine the two known front pieces first, $\frac14+\frac15=\frac{9}{20}$, then apply the outer $\frac12$.

$\frac12\cdot\frac{9}{20} = \frac{9}{40}, \qquad \text{so} \qquad \sum_{k=4}^{n}\frac{1}{k(k+2)} = \frac{9}{40}-\frac12\left(\frac1{n+1}+\frac1{n+2}\right).$

In a gap-1 telescope, the negative piece from bracket $k$ is $-\frac1{k+1}$, and the very next bracket, $k+1$, contributes the positive piece $+\frac1{k+1}$: cancellation is immediate, one bracket apart, so only ONE piece survives at each end.

In a gap-2 telescope, the negative piece from bracket $k$ is $-\frac1{k+2}$, and that value does not reappear with a plus sign until bracket $k+2$, which is TWO brackets later, not the very next one:

$\text{bracket } k:\ -\frac1{k+2} \qquad\qquad \text{bracket } k+2:\ +\frac1{k+2}.$

Every bracket's negative piece is waiting two steps for its partner, which means the first TWO positive pieces at the start of the range, and the last TWO negative pieces at the end, never reach far enough to meet a partner within the sum. Whatever the specific numbers are, this is a structural fact about gap-2 splits: the size of the gap between the two factors sets how many brackets apart a term meets its partner, and that count is exactly how many survivors pile up at each end.

Multiply top and bottom by the conjugate $\sqrt{k+1}-\sqrt{k}$, which turns the denominator into a difference of squares.

$\frac{1}{\sqrt{k}+\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)} = \frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k} = \sqrt{k+1}-\sqrt{k}.$

The denominator reduces to $1$, so the whole fraction is just the difference of consecutive square roots.

By part A, every term equals $\sqrt{k+1}-\sqrt{k}$, the mirror telescoping form with $b_k=\sqrt{k}$, which collapses to $\sqrt{n+1}-1$.

$\sum_{k=1}^{143}\frac{1}{\sqrt{k}+\sqrt{k+1}} = \sqrt{144}-1 = 12-1 = 11.$

The finite closed form for this sum, for any upper limit $n$, is $\sqrt{n+1}-1$, exactly as in part A. An infinite telescoping sum settles precisely when its surviving tail term shrinks to zero as $n$ grows, and here the surviving tail is $\sqrt{n+1}$ itself, which climbs without any bound rather than shrinking.

$\sqrt{n+1}-1 \;\longrightarrow\; \infty \quad \text{as} \quad n\to\infty.$

So unlike the workhorse fraction $\frac1{k(k+1)}$, whose tail $\frac1{n+1}$ does shrink to zero, this square-root sum has no finite total, no matter how many terms are added. Rationalizing the denominator was necessary to see the telescope at all, but it does not change what decides convergence: only the behavior of the surviving tail term does.