Finding Telescoping Patterns

Finding the Difference Hidden in a Term

A telescoping series is useful only after its cancellation pattern has been found. In the previous tutorial, we saw how a representation $a_n=b_n-b_{n+1}$ reduces a partial sum to its boundary terms. The practical question now is how to discover such a representation from a formula for $a_n$.

Several algebraic structures often signal a useful difference: a denominator that factors into nearby linear terms, a radical sum that can be rationalized, or factorials with consecutive indices. The goal is not to guess cancellation from the first few terms. Instead, rewrite the general term exactly, verify the identity algebraically, and then calculate a finite partial sum with its actual endpoints.

Notation: When a sequence $(b_n)$ is in view, write $\Delta b_n=b_n-b_{n+1}$. Finding a one-step telescoping pattern means finding $b_n$ for which $a_n=\Delta b_n$. A fixed shift $k$ instead gives terms of the form $b_n-b_{n+k}$; in that case, as many as $k$ terms can remain at each end of a finite sum.

For rational expressions, factoring and partial fractions are natural first steps. A difference of reciprocals has a numerator equal to the difference of its denominators:

\frac{1}{n+p}-\frac{1}{n+q} =\frac{q-p}{(n+p)(n+q)}.

Thus a product of two distinct linear factors can produce a difference of shifted reciprocals, provided the numerator is a suitable constant. For radicals, multiplying by a conjugate can turn a reciprocal into a difference. For factorials, writing both terms with the same denominator often reveals consecutive terms that cancel.

Boundary Terms for a Fixed Shift

The one-step identity from the previous tutorial has a useful generalization. With a shift of $k$, the positive and negative copies of the sequence overlap except near the two ends. The following formula records those ends and also covers short sums, where the positive and negative ranges do not overlap.

Theorem (Finite Telescoping Identity for a Fixed Shift): Let $(b_n)$ be a sequence, and let $k,m,N$ be positive integers with $m\leq N$. Then $$ \sum_{n=m}^{N}(b_n-b_{n+k}) = \sum_{j=m}^{\min(N,m+k-1)}b_j - \sum_{j=\max(m+k,N+1)}^{N+k}b_j. $$

Proof. In the expanded sum, the positive terms have indices $m,m+1,\ldots,N$, while the negative terms have indices $m+k,m+k+1,\ldots,N+k$. If $N\geq m+k$, the indices from $m+k$ through $N$ occur once with each sign and cancel. The uncancelled positive indices are then $m$ through $m+k-1$, and the uncancelled negative indices are $N+1$ through $N+k$, exactly as in the formula.

If $N<m+k$, the two index ranges do not overlap: the positive indices end at $N$, before the negative indices begin at $m+k$. The formula then gives the positive sum from $m$ through $N$, because $\min(N,m+k-1)=N$, and the negative sum from $m+k$ through $N+k$, because $\max(m+k,N+1)=m+k$. This is precisely the expanded sum in this case as well. The boundary cases are therefore included. $\square$

In particular, if $N\geq k$ and the sum begins at $m=1$, this becomes

\sum_{n=1}^{N}(b_n-b_{n+k}) = \sum_{j=1}^{k}b_j-\sum_{j=N+1}^{N+k}b_j.

This version makes the pattern especially visible: a fixed shift leaves the first $k$ terms of the sequence at the beginning and $k$ terms at the far end. If the final terms tend to zero as $N$ grows, only the initial block contributes to the limit. The condition $N\geq k$ matters for this simplified form; the general theorem handles smaller partial sums without pretending that the two boundary blocks are already separated.

Factoring and Partial Fractions

The next theorem turns a common algebraic pattern into a reusable series formula. It is a direct application of partial fractions and the fixed-shift identity, but stating the result explicitly makes the surviving finite block easy to identify.

Theorem (Sum of a Rational Fixed-Shift Telescope): Let $c\in\mathbb{R}$, and let $p,q$ be positive integers with $p<q$. Then $$ \sum_{n=1}^{\infty}\frac{c}{(n+p)(n+q)} = \frac{c}{q-p}\sum_{j=p+1}^{q}\frac{1}{j}. $$

Proof. Since $q-p>0$, direct subtraction gives, for each $n\geq1$,

\frac{1}{n+p}-\frac{1}{n+q} = \frac{(n+q)-(n+p)}{(n+p)(n+q)} = \frac{q-p}{(n+p)(n+q)}.

Therefore

\frac{c}{(n+p)(n+q)} = \frac{c}{q-p} \left(\frac{1}{n+p}-\frac{1}{n+q}\right).

Put $k=q-p$ and $b_n=1/(n+p)$. The term in parentheses is $b_n-b_{n+k}$. For $N\geq k$, the fixed-shift identity gives

\begin{aligned} \sum_{n=1}^{N}\frac{c}{(n+p)(n+q)} &=\frac{c}{q-p} \left(\sum_{j=1}^{k}\frac{1}{j+p} -\sum_{j=N+1}^{N+k}\frac{1}{j+p}\right)\\ &=\frac{c}{q-p} \left(\sum_{j=p+1}^{q}\frac{1}{j} -\sum_{j=N+1}^{N+k}\frac{1}{j+p}\right). \end{aligned}

For each $j$ in the final sum, $j+p\geq N+1+p$, so \[ 0\leq \sum_{j=N+1}^{N+k}\frac{1}{j+p} \leq \frac{k}{N+1+p}. \] The right-hand side tends to zero. Hence the partial sums converge to the stated value. This proof applies when $c$ is positive, zero, or negative, since the vanishing boundary sum is multiplied by the fixed constant $c/(q-p)$. $\square$

Worked Example: Factoring a Quadratic Denominator

Consider $\sum_{n=1}^{\infty}6/((n+2)(n+5))$. The factors differ by three, and the reciprocal-difference identity gives

\frac{6}{(n+2)(n+5)} = 2\left(\frac{1}{n+2}-\frac{1}{n+5}\right),

because the difference in parentheses has numerator $(n+5)-(n+2)=3$. For $N\geq3$, expanding and cancelling yields

\begin{aligned} S_N &=2\sum_{n=1}^{N}\left(\frac{1}{n+2}-\frac{1}{n+5}\right)\\ &=2\left(\frac13+\frac14+\frac15-\frac{1}{N+3} -\frac{1}{N+4}-\frac{1}{N+5}\right). \end{aligned}

The initial block is $1/3+1/4+1/5=20/60+15/60+12/60=47/60$. The three final reciprocals tend to zero, so \[ \sum_{n=1}^{\infty}\frac{6}{(n+2)(n+5)} =2\cdot\frac{47}{60} =\frac{47}{30}. \] The factorization identifies not just that cancellation occurs, but exactly how many initial terms survive: three, because the shifts differ by three.

Rationalizing a Radical Denominator

A denominator containing a sum of square roots may conceal a difference rather than a difference of reciprocals. The conjugate identity \[ (\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})=x-y \] can expose it. The sign and the constant produced by $x-y$ must be checked; the conjugate is an algebraic tool, not a reason to assume convergence.

Worked Example: Conjugates Reveal a Two-Step Pattern

For $n\geq1$, the denominator $\sqrt{n+3}+\sqrt{n+1}$ is positive. Multiplying its reciprocal by the conjugate gives

\begin{aligned} \frac{1}{\sqrt{n+3}+\sqrt{n+1}} &=\frac{\sqrt{n+3}-\sqrt{n+1}} {(\sqrt{n+3}+\sqrt{n+1})(\sqrt{n+3}-\sqrt{n+1})}\\ &=\frac{\sqrt{n+3}-\sqrt{n+1}}{(n+3)-(n+1)}\\ &=\frac12\bigl(\sqrt{n+3}-\sqrt{n+1}\bigr). \end{aligned}

The shift is two. Summing the finite differences, the first few positive terms are $\sqrt4,\sqrt5,\ldots,\sqrt{N+3}$, and the negative terms are $\sqrt2,\sqrt3,\ldots,\sqrt{N+1}$. Their overlap cancels, leaving

S_N =\frac12\left(\sqrt{N+2}+\sqrt{N+3}-\sqrt2-\sqrt3\right).

This formula also holds for $N=1$: its right-hand side is \[ \frac12(\sqrt3+\sqrt4-\sqrt2-\sqrt3) =\frac12(2-\sqrt2), \] which equals the original first term $1/(\sqrt4+\sqrt2)$. As $N$ increases, both $\sqrt{N+2}$ and $\sqrt{N+3}$ grow without bound, so $S_N$ does not have a finite limit. The series diverges even though the terms have an exact telescoping representation. The boundary terms decide the outcome.

Matching Consecutive Factorials

Factorials provide another common source of hidden differences. When two factorials have neighboring indices, rewrite them over the larger factorial. The resulting numerator may be a simple factor of $n$, and the expression can become a difference of consecutive reciprocal factorials.

Worked Example: A Factorial Difference with Sum One

For $n\geq1$, the factorial identity $(n+1)!=(n+1)n!$ gives

\begin{aligned} \frac{n}{(n+1)!} &=\frac{n+1-1}{(n+1)!}\\ &=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}\\ &=\frac{1}{n!}-\frac{1}{(n+1)!}. \end{aligned}

Thus the $N$th partial sum is

\begin{aligned} S_N &=\sum_{n=1}^{N}\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right)\\ &=1-\frac{1}{(N+1)!}. \end{aligned}

Since $(N+1)!\to\infty$, its reciprocal tends to zero. Therefore \[ \sum_{n=1}^{\infty}\frac{n}{(n+1)!}=1. \] The decisive step was putting the two terms over a shared factorial and checking the numerator $n+1-1=n$.

A Reliable Search Procedure

The same sequence of checks works across these different algebraic forms. A promising rewrite is only the beginning: its identity, shift, endpoints, and limiting boundary terms all need to be verified.

Inspect the structure.
Factor polynomial denominators, compare nearby indices, or look for conjugates and consecutive factorials.

Propose a difference.
Choose a candidate $b_n$, or a fixed shift $k$, that matches the visible factors or indices.

Verify the algebra.
Combine the proposed terms over a common denominator, or multiply by the conjugate, and check that the result is exactly $a_n$.

Calculate a finite sum.
Write down the surviving initial and final terms, including any overlap for small $N$.

Take the boundary limit.
Determine whether the remaining terms have a finite limit before assigning a sum to the infinite series.

Partial fractions can also warn that a proposed telescope is not present in the simplest form. For example, decomposing a rational term may produce two reciprocal terms with coefficients of the same sign rather than opposite signs. That decomposition does not give the immediate cancellation $b_n-b_{n+k}$. It may still be possible to use a different representation, but cancellation should not be claimed until an exact difference identity has been established.

A frequent error is to see matching expressions at nearby indices and cancel them informally across an infinite series. Cancellation is justified first in each finite partial sum. The Finite Telescoping Identity for a Fixed Shift makes the boundary terms explicit, while the earlier Convergence Criterion for a Telescoping Series explains why the boundary behavior determines convergence in the one-step case.

Takeaway: Factor, decompose, rationalize, or rewrite factorials to search for a difference. Then verify the identity and compute a finite partial sum: the surviving boundary terms, not the visual appearance of cancellation, determine whether the series converges and what its sum is.

Check Your Understanding

Use an exact finite-sum identity to answer each question.

For a shift of $k$, which positive and negative terms remain when $N\geq k$ in $\sum_{n=1}^{N}(b_n-b_{n+k})$?
Rewrite $4/((n+1)(n+4))$ as a difference of shifted reciprocals, then find the sum of the series beginning at $n=1$.
What conjugate should be used to rewrite $1/(\sqrt{n+5}+\sqrt{n+2})$, and what constant appears in the denominator after rationalization?
Verify the difference representation of $n/(n+1)!$ by combining its two terms over a common denominator.
Why does finding a telescoping representation not, by itself, establish convergence?

What You'll Learn