Why a Fixed Index Gap Telescopes
A term such as \(1/[n(n+2)]\) contains two factors whose indices differ by a fixed amount. Partial fractions turn that product into a difference of two reciprocals. When these differences are added over consecutive values of \(n\), many terms cancel.
The cancellation can be less straightforward when the gap is larger than the number of terms being added. In that case, the two shifted index ranges do not overlap, even though the boundary blocks in a useful formula can overlap with each other. Keeping those situations distinct gives a formula that works for every finite sum.
Decomposing the Reciprocal Product
Let \(k\) be a positive integer and let \(n\) be a positive integer. The fixed-gap reciprocal product has the decomposition
To verify it, combine the two fractions on the right:
The factor \(1/k\) is essential. It compensates for the \(k\) in the numerator obtained by subtracting the reciprocals.
The Shifted-Sum Identity
Apply the decomposition separately to every summand. The sum of the first reciprocals has indices from \(m\) through \(N\). In the second sum, the index \(n+k\) runs from \(m+k\) through \(N+k\). Therefore
This identity already exposes the cancellation: any reciprocal whose index occurs in both sums appears once with a positive sign and once with a negative sign. But whether the index ranges overlap depends on the number \(L\) of terms compared with the gap \(k\).
The Boundary-Block Formula
For all positive integers \(m\leq N\) and \(k\), the shifted-sum identity can be written using two blocks of \(k\) reciprocals:
The first block begins at the starting index \(m\). The second begins just after the ending index \(N\), and it is subtracted. These are the surviving boundary blocks. The formula is valid even if the two blocks overlap; in that case their shared terms cancel against each other.
Here is a complete justification that covers all possible lengths.
The original ranges \([m,N]\) and \([m+k,N+k]\) overlap from \(m+k\) through \(N\). Each reciprocal in that common range occurs with opposite signs and cancels. The uncancelled positive indices are \(m,\ldots,m+k-1\); the uncancelled negative indices are \(N+1,\ldots,N+k\).
The original ranges are \([m,m+k-1]\) and \([m+k,m+2k-1]\). They are adjacent, not overlapping. They already are the first and last boundary blocks in the formula, since \(N=m+k-1\) and \(N+1=m+k\).
The original ranges are disjoint, so no cancellation occurs between those two ranges. Instead, the boundary blocks overlap: the first runs from \(m\) to \(m+k-1\), and the second from \(N+1=m+L\) to \(N+k=m+L+k-1\). Their shared indices \(m+L,\ldots,m+k-1\) cancel. What remains is the positive block \(m,\ldots,N\) and the negative block \(m+k,\ldots,N+k\), exactly as in the shifted-sum identity.
This case-by-case argument shows why it is important not to claim that the original index ranges always overlap. The boundary-block formula is universal, but the location of the cancellation depends on the relative sizes of \(L\) and \(k\).
Worked Examples
Worked Example: A Sum Longer Than Its Gap
Evaluate \(\displaystyle\sum_{n=1}^{5}\frac{1}{n(n+2)}\). Here \(m=1\), \(N=5\), and \(k=2\), so \(L=5>2\). The original shifted ranges overlap.
First use the decomposition and write out the terms:
The reciprocals \(1/3\), \(1/4\), and \(1/5\) cancel. Now evaluate the remaining boundary terms:
Multiplying by \(1/2\) gives \(25/42\). Check directly using the original summands:
The direct sum agrees with the boundary calculation.
Worked Example: One Term and a Larger Gap
Evaluate \(\displaystyle\sum_{n=1}^{1}\frac{1}{n(n+3)}\). Here \(m=N=1\), \(k=3\), and \(L=1<3\). The original index ranges are \([1,1]\) and \([4,4]\), which are disjoint.
The boundary-block formula gives
In this calculation, \(1/2\) and \(1/3\) cancel because they occur in the two boundary blocks, not because the original index ranges overlap. A direct check gives \(1/[1(1+3)]=1/4\), as required.
Worked Example: A Sum Exactly as Long as Its Gap
Evaluate \(\displaystyle\sum_{n=2}^{3}\frac{1}{n(n+2)}\). In this case \(m=2\), \(N=3\), \(k=2\), and \(L=2=k\). The original ranges are adjacent.
Using the boundary blocks:
Check the original terms: \(1/[2(4)]=1/8\) and \(1/[3(5)]=1/15\). Their sum is \(1/8+1/15=15/120+8/120=23/120\), which agrees.
Worked Example: A Longer Sum Starting Away from One
Evaluate \(\displaystyle\sum_{n=2}^{7}\frac{1}{n(n+2)}\). Here \(m=2\), \(N=7\), and \(k=2\). Since \(L=6>2\), the original ranges overlap.
The boundary blocks are the first two reciprocals starting at \(2\), and the last two reciprocals starting at \(8\):
For a direct check, the six original terms are \(1/8,1/15,1/24,1/35,1/48,1/63\). With common denominator \(5040\), their sum is
The direct calculation confirms the boundary result.
A Common Mistake: Assuming the Original Ranges Always Overlap
A tempting but incorrect explanation is that every finite sum cancels its middle terms in the original two shifted sums. That claim fails when \(L\leq k\). For example, with \(m=N=1\) and \(k=3\), the original ranges are \([1,1]\) and \([4,4]\). There is no shared index to cancel. The shifted-sum difference is simply \(1-1/4\), and the factor \(1/3\) gives \(1/4\).
The boundary blocks in this same example are different: they run from \(1\) to \(3\) and from \(2\) to \(4\). Their shared indices, \(2\) and \(3\), cancel, leaving \(1-1/4\). Thus, the final boundary formula remains correct, but the explanation must identify which two ranges actually overlap.
Another error is to omit the factor \(1/k\) in the partial-fraction decomposition. For \(k=2\), the correct identity is
not just the difference in parentheses. Omitting \(1/2\) doubles every term and therefore doubles the value of every partial sum.
Why the Boundary View Is Useful
A long finite sum may contain many terms, but its telescoping form can depend on only a few reciprocals at the two ends. The fixed gap \(k\) determines how many terms appear in each boundary block. This makes the finite sum easier to compute and clarifies exactly where cancellation happens.
The boundary formula also prevents a misleading picture of cancellation. If there are more than \(k\) terms, the original shifted sums share indices. If there are fewer than \(k\) terms, they do not; instead, the two boundary blocks overlap. Distinguishing these cases is necessary for a correct finite-sum argument. The formula describes the finite partial sum itself, without requiring any assumption about whether an associated infinite series converges.
Check Your Understanding
Use the fixed-gap decomposition and boundary blocks to answer these questions.
- For \(k=4\), write \(1/[n(n+4)]\) as a difference of reciprocals multiplied by a constant.
- For \(m=3\), \(N=7\), and \(k=2\), do the original shifted index ranges overlap, touch, or remain disjoint?
- Write the boundary-block formula for \(\displaystyle\sum_{n=2}^{4}1/[n(n+3)]\). Do the boundary blocks overlap?
- For a sum with \(L<k\), which ranges are disjoint, and which ranges may overlap and cancel?
- Explain why the factor \(1/k\) is required in the reciprocal-product decomposition.
