Why Consecutive Differences Simplify a Sum
A long sum can be much easier to evaluate when each term is the difference between two consecutive values of a sequence. In the expanded sum, most values appear twice: once with a positive sign and once with a negative sign. Those matching terms cancel, leaving only the values at the two ends.
This cancellation pattern is called telescoping. The name describes how the middle of a sum seems to fold away, exposing its boundary terms. The essential work is to write the terms in the correct order and keep track of the signs and indices.
The Basic Finite-Sum Formula
Suppose \(b_k\) is defined for the indices from \(m\) through \(N+1\), where \(m\leq N\), and suppose \(a_k=b_k-b_{k+1}\) for \(m\leq k\leq N\). Expanding the sum gives
Each value from \(b_{m+1}\) to \(b_N\) occurs once positively and once negatively. Only the first positive value, \(b_m\), and the final negative value, \(-b_{N+1}\), remain.
The upper boundary is \(b_{N+1}\), not \(b_N\), because the last summand is \(b_N-b_{N+1}\). This is an important consequence of the index shift in the second part of each difference.
Why the Formula Works
The cancellation can be justified by splitting the finite sum into two sums and shifting the index in the second one. Since both sums contain finitely many terms, this rearrangement is valid.
Write the sum of differences as the sum of the \(b_k\) values minus the sum of the \(b_{k+1}\) values.
As \(k\) runs from \(m\) to \(N\), the index \(k+1\) runs from \(m+1\) to \(N+1\).
The two sums have every term from \(b_{m+1}\) through \(b_N\) in common, with opposite signs.
In equations, the derivation is
This proves the finite telescoping formula. It also explains why simply looking at the first few summands can be useful: their signs reveal the cancellation pattern. The formula, however, follows from the complete finite expansion, not from guessing which terms might disappear.
Worked Example: Differences of Consecutive Squares
Worked Example: Differences of Consecutive Squares
Find the finite sum \(\sum_{k=1}^{4}((k+1)^2-k^2)\) by displaying its cancellation.
Here, the consecutive sequence values are \(b_k=k^2\). The four terms are
Grouping the positive and negative values shows what cancels:
The same result follows from the boundary formula: \(b_1=1^2=1\) and \(b_5=5^2=25\), so the sum is \(b_1-b_5=1-25\) if the terms were written \(b_k-b_{k+1}\). In this example the displayed summands are instead \((k+1)^2-k^2=b_{k+1}-b_k\), so the boundary order is reversed: \(b_5-b_1=25-1=24\).
Check: Evaluating the four differences individually gives \(3+5+7+9=24\), which agrees with the cancellation result.
This example illustrates a sign choice that matters. The formula \(b_m-b_{N+1}\) applies to differences written \(b_k-b_{k+1}\). If each difference is written in the opposite order, \(b_{k+1}-b_k\), the surviving boundary terms are also in the opposite order.
Worked Example: Differences of Sines
Worked Example: Differences of Sines
For a positive integer \(N\), simplify the finite sum
Set \(b_k=\sin(1/k)\). Then the summand is \(b_k-b_{k+1}\), so the formula gives
To see the cancellation directly, the first few terms are
Every intermediate sine value occurs once with each sign. For a direct check, if \(N=3\), expansion gives \((\sin 1-\sin(1/2))+(\sin(1/2)-\sin(1/3))+(\sin(1/3)-\sin(1/4))=\sin 1-\sin(1/4)\), exactly the formula with \(N=3\).
Worked Example: A Periodic Sequence of Boundary Values
Worked Example: A Periodic Sequence of Boundary Values
Evaluate \(\sum_{k=1}^{5}\left(\cos(k\pi/3)-\cos((k+1)\pi/3)\right)\).
Take \(b_k=\cos(k\pi/3)\). The needed sequence values, from \(b_1\) through \(b_6\), are
Writing the five differences shows every intermediate value cancel:
As a check, the five differences are \(1,\frac12,-\frac12,-1,-\frac12\). Their sum is \(1+\frac12-\frac12-1-\frac12=-\frac12\), agreeing with the boundary calculation.
Signs, Indices, and What Remains
When identifying a telescoping pattern, write down the sequence of signed terms before simplifying. The sign order determines which boundary values survive. In particular, the two common forms have opposite results:
| Summand | Finite sum from \(m\) to \(N\) |
|---|---|
| \(b_k-b_{k+1}\) | \(b_m-b_{N+1}\) |
| \(b_{k+1}-b_k\) | \(b_{N+1}-b_m\) |
For the second row, each summand is the negative of the first form. Equivalently, applying the first formula and multiplying the result by \(-1\) gives \(b_{N+1}-b_m\). Checking the first and last summands is a reliable way to confirm the boundary indices: the first term contains \(b_m\) and \(b_{m+1}\), while the last contains \(b_N\) and \(b_{N+1}\).
A Common Mistake: Keeping the Wrong Endpoint
A frequent error is to stop the boundary term at \(b_N\). Consider the sum \(\sum_{k=2}^{4}(b_k-b_{k+1})\). A wrong calculation might claim that the result is \(b_2-b_4\), as though the last summand ended at \(b_4\). But the actual expansion is
The last summand is \(b_4-b_5\), so \(b_5\), not \(b_4\), is the final boundary value. The correction is to inspect the index \(k+1\) in the second part of every summand: at the final index \(k=4\), it becomes \(4+1=5\).
A second version of the same mistake is to use the correct endpoints but reverse their signs. For \(b_k-b_{k+1}\), the result is \(b_m-b_{N+1}\); for \(b_{k+1}-b_k\), it is \(b_{N+1}-b_m\). Expanding the first and last terms usually exposes a sign error before it spreads through the calculation.
Finite Cancellation and Infinite Series
A telescoping expression for each term provides a simple formula for every finite partial sum. For example, the first \(N\) terms of a series with \(a_k=b_k-b_{k+1}\) have sum \(b_1-b_{N+1}\). This finite identity is exact for each permitted \(N\).
It is important not to confuse that identity with an evaluation of an infinite series. To decide whether an infinite series converges, one must study what happens to its partial sums as \(N\) increases. Telescoping identifies those partial sums and their boundary terms; it does not, by cancellation alone, establish a convergence conclusion. The finite calculation is the necessary first step.
Check Your Understanding
Use the cancellation pattern and the index range carefully in each question.
- Write out \(\sum_{k=3}^{6}(b_k-b_{k+1})\). Which terms remain after cancellation?
- For \(\sum_{k=1}^{N}(c_{k+1}-c_k)\), what is the result in terms of the boundary values?
- In \(\sum_{k=2}^{5}(d_k-d_{k+1})\), why is the final boundary term \(d_6\), not \(d_5\)?
- Evaluate \(\sum_{k=1}^{4}\left(\frac{1}{2^k}-\frac{1}{2^{k+1}}\right)\) by identifying the sequence values at the boundaries.
- Does the fact that a finite sum telescopes by itself prove that the corresponding infinite series converges? Explain briefly.
