Finding a Telescoping Decomposition

Why Finding the Constants Matters

A series may look difficult to sum because its terms are written as a single fraction. If the fraction can be rewritten as a difference of shifted reciprocals, many terms cancel when we form partial sums. The challenge is to find the constants in that rewrite accurately: a wrong sign can destroy the cancellation and give an incorrect sum.

The reliable approach is to clear denominators. This turns the proposed decomposition into an identity between polynomials. Matching coefficients then determines the constants, and substituting them back into the identity checks the result. Once the decomposition is verified, partial sums reveal exactly what remains after cancellation.

Definition: A telescoping decomposition rewrites a term as a sum or difference of simpler terms arranged so that terms in consecutive summands cancel in a partial sum. For instance, \(A\left(\frac{1}{n+a}-\frac{1}{n+b}\right)\) is a reciprocal-difference form. Whether it telescopes usefully depends on the shifts and the range of summation.

Finding Constants by Clearing Denominators

For a fraction with two distinct linear factors in its denominator, a common proposed form is

\frac{C}{(n+a)(n+b)} = \frac{A}{n+a}+\frac{B}{n+b}.

To determine \(A\) and \(B\), multiply both sides by \((n+a)(n+b)\). The proposed identity is equivalent to

$$ C=A(n+b)+B(n+a). $$

Now both sides are polynomials in \(n\). Their coefficients must agree. In particular, the coefficient of \(n\) on the right is \(A+B\), and its constant term is \(bA+aB\). If \(a\ne b\), these two equations determine \(A\) and \(B\). The decomposition becomes a reciprocal difference precisely when \(A+B=0\), so that \(B=-A\).

This method works because multiplying by the common denominator preserves equality wherever the original denominator is nonzero. Conversely, if the resulting polynomial identity holds, division by that nonzero denominator recovers the original rational identity. Thus clearing denominators both provides a way to solve for the constants and a direct way to verify them.

Worked Examples

Worked Example: Find and use a two-term reciprocal difference

Evaluate \(\displaystyle\sum_{n=1}^{\infty}\frac{3}{(n+1)(n+4)}\), if it converges, by determining the constants in a reciprocal-difference decomposition.

Begin by proposing

\frac{3}{(n+1)(n+4)} = \frac{A}{n+1}+\frac{B}{n+4}.

Clear the denominator \((n+1)(n+4)\). This gives the polynomial identity

$$ 3=A(n+4)+B(n+1) =(A+B)n+(4A+B). $$

The coefficient of \(n\) on the left is \(0\), so \(A+B=0\). The constant terms give \(4A+B=3\). Since \(B=-A\), the second equation becomes \(4A-A=3\), hence \(3A=3\), \(A=1\), and \(B=-1\). Therefore

\frac{3}{(n+1)(n+4)} = \frac{1}{n+1}-\frac{1}{n+4}.

Check by recombining the fractions:

\frac{1}{n+1}-\frac{1}{n+4} = \frac{(n+4)-(n+1)}{(n+1)(n+4)} = \frac{3}{(n+1)(n+4)}.

For a partial sum with \(N\ge 4\), write out the shifted ranges:

\begin{aligned} S_N &=\sum_{n=1}^{N}\left(\frac{1}{n+1}-\frac{1}{n+4}\right)\\ &=\left(\frac12+\frac13+\frac14+\frac15+\cdots+\frac{1}{N+1}\right) -\left(\frac15+\frac16+\cdots+\frac{1}{N+4}\right)\\ &=\frac12+\frac13+\frac14-\frac{1}{N+2}-\frac{1}{N+3}-\frac{1}{N+4}. \end{aligned}

The common terms cancel, leaving three initial terms and three final terms. As \(N\) increases, each of the three final terms tends to \(0\). Therefore

\sum_{n=1}^{\infty}\frac{3}{(n+1)(n+4)} = \frac12+\frac13+\frac14 = \frac{6+4+3}{12} = \frac{13}{12}.

As a check on the cancellation, take \(N=4\). The partial sum is \(1/2-1/5+1/3-1/6+1/4-1/7+1/5-1/8\). The \(1/5\) terms cancel, and the result is \(13/12-1/6-1/7-1/8\), which agrees with the displayed formula for \(S_N\).

Worked Example: Determine the constants and evaluate the series

Evaluate \(\displaystyle\sum_{n=1}^{\infty}\frac{6}{(n+1)(n+3)}\). Find the constants rather than guessing the decomposition.

Set

\frac{6}{(n+1)(n+3)} = \frac{A}{n+1}+\frac{B}{n+3}.

After clearing denominators,

$$ 6=A(n+3)+B(n+1) =(A+B)n+(3A+B). $$

Matching coefficients gives \(A+B=0\) and \(3A+B=6\). Substitute \(B=-A\) into the second equation: \(3A-A=6\), so \(2A=6\), \(A=3\), and \(B=-3\). Thus

\frac{6}{(n+1)(n+3)} = \frac{3}{n+1}-\frac{3}{n+3}.

Recombining verifies the identity:

\frac{3}{n+1}-\frac{3}{n+3} = \frac{3(n+3)-3(n+1)}{(n+1)(n+3)} = \frac{6}{(n+1)(n+3)}.

For \(N\ge2\), the partial sum is

\begin{aligned} S_N &=3\left(\frac12+\frac13+\frac14+\cdots+\frac{1}{N+1}\right) -3\left(\frac14+\frac15+\cdots+\frac{1}{N+3}\right)\\ &=\frac32+1-\frac{3}{N+2}-\frac{3}{N+3}\\ &=\frac52-\frac{3}{N+2}-\frac{3}{N+3}. \end{aligned}

The terms from \(1/4\) through \(1/(N+1)\) cancel. The two remaining terms at the upper end tend to \(0\), so the series converges to

\lim_{N\to\infty}S_N=\frac52.

For a numerical check at \(N=2\), the original terms are \(6/(2\cdot4)=3/4\) and \(6/(3\cdot5)=2/5\), giving \(3/4+2/5=15/20+8/20=23/20\). The partial-sum formula gives \(5/2-3/4-3/5=50/20-15/20-12/20=23/20\), as required.

Worked Example: Solve for three constants in a telescoping decomposition

Find constants \(A,B,C\) such that

\frac{2}{(n+1)(n+2)(n+3)} = \frac{A}{n+1}+\frac{B}{n+2}+\frac{C}{n+3},

and use the decomposition to evaluate the series starting at \(n=1\).

Clear all three denominator factors:

$$ 2=A(n+2)(n+3)+B(n+1)(n+3)+C(n+1)(n+2). $$

Expand each product and collect powers of \(n\):

\begin{aligned} 2 &=A(n^2+5n+6)+B(n^2+4n+3)+C(n^2+3n+2)\\ &=(A+B+C)n^2+(5A+4B+3C)n+(6A+3B+2C). \end{aligned}

The left side has zero coefficients for \(n^2\) and \(n\), and constant term \(2\). Hence

A+B+C=0,\qquad 5A+4B+3C=0,\qquad 6A+3B+2C=2.

Subtract the first equation from the second to obtain \(4A+3B+2C=0\). Subtract this equation from the third: \(2A=2\), so \(A=1\). The first equation now gives \(B+C=-1\), and \(4A+3B+2C=0\) becomes \(4+3B+2C=0\). Substituting \(C=-1-B\) gives \(4+3B+2(-1-B)=0\), or \(2+B=0\). Thus \(B=-2\) and \(C=1\). The decomposition is

\frac{2}{(n+1)(n+2)(n+3)} = \frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}.

To verify it, put the right side over a common denominator. Its numerator is

$$ (n+2)(n+3)-2(n+1)(n+3)+(n+1)(n+2)=2. $$

For \(N\ge2\), summing the three shifted terms makes the interior terms cancel. The first two lower-end terms and the last two upper-end terms give

\begin{aligned} S_N &=\sum_{n=1}^{N}\left(\frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}\right)\\ &=\frac12-\frac13-\frac{1}{N+2}+\frac{1}{N+3}\\ &=\frac16-\frac{1}{N+2}+\frac{1}{N+3}. \end{aligned}

The last two fractions tend to \(0\). Therefore the infinite series sums to \(1/6\). Check the finite formula at \(N=2\): the original terms are \(2/(2\cdot3\cdot4)=1/12\) and \(2/(3\cdot4\cdot5)=1/30\), whose sum is \(5/60+2/60=7/60\). The formula gives \(1/6-1/4+1/5=10/60-15/60+12/60=7/60\).

Why the Partial-Sum Check Is Essential

A decomposition does not by itself give the value of an infinite series. The sum is defined through the limit of its partial sums: if \(S_N\) is the sum of the first \(N\) terms and \(\lim_{N\to\infty}S_N=L\), then the series converges to \(L\). In the examples, cancellation produces a formula for \(S_N\), and the remaining terms are explicitly shown to tend to zero. This is the step that justifies passing from a finite cancellation pattern to an infinite sum.

The shifts determine what survives. For example, in \(1/(n+1)-1/(n+4)\), the second reciprocal begins three places later than the first. In a partial sum, all overlapping terms cancel, leaving three terms at each end. The upper-end terms vanish in the limit, while the lower-end terms determine the value. When there are more reciprocal terms, as in the third example, cancellation may involve several contributions at each index; writing the first few and last few terms prevents mistakes about what remains.

Common Sign Error

A frequent mistake is to reverse the constants in the two-factor decomposition. For the fraction in the first example, an incorrect choice would be

\frac{3}{(n+1)(n+4)} \overset{\text{wrong}}{=} -\frac{1}{n+1}+\frac{1}{n+4}.

Recombining the right side exposes the error:

-\frac{1}{n+1}+\frac{1}{n+4} = \frac{-(n+4)+(n+1)}{(n+1)(n+4)} = -\frac{3}{(n+1)(n+4)}.

The numerator is \(-3\), not \(3\). The correction is to use \(A=1\) and \(B=-1\), so the numerator becomes \((n+4)-(n+1)=3\). Always verify a decomposition by recombining its terms before using it in a sum.

A Reliable Procedure

Choose a proposed form.
Use one reciprocal term for each distinct linear denominator factor in the decomposition being sought.

Clear denominators.
Multiply through by the common denominator, then expand and collect powers of the index.

Determine the constants.
Equate coefficients on both sides and solve the resulting equations.

Verify the identity.
Substitute the constants into the cleared-denominator equation or recombine the fractions.

Sum and take a limit when needed.
Write a partial sum, identify the uncancelled terms, and evaluate their limit before stating an infinite sum.

Check Your Understanding

For each question, use denominator clearing and show enough work to verify the resulting decomposition.

Find \(A\) and \(B\) such that \(\frac{2}{(n+2)(n+5)}=\frac{A}{n+2}+\frac{B}{n+5}\). Which reciprocal-difference form does this give?
For the decomposition in Question 1, write the first four terms of the partial sum from \(n=1\). Identify the terms that cancel.
Find constants \(A,B,C\) for \(\frac{2}{(n+1)(n+2)(n+3)}=\frac{A}{n+1}+\frac{B}{n+2}+\frac{C}{n+3}\).
Why is it necessary to verify the decomposition before using it to evaluate a series?
When a partial sum has been simplified, what additional step is required before concluding that an infinite series has a particular sum?