Telescoping Rational Terms by Partial Fractions

Why Partial Fractions Reveal Cancellation

A rational term can conceal a simple difference. For instance, the denominator \(n(n+1)\) looks like a product, but its reciprocal can be rewritten as the difference of two reciprocals. Summing those differences makes most terms cancel, leaving only a few terms at the ends of a finite sum.

This is the connection between partial fractions and telescoping: partial fractions rewrite an individual term; telescoping describes what happens when those rewritten terms are added. The cancellation pattern is useful for finding partial sums and, when their limit is finite, the sum of an infinite series.

Definition: For a series \(\sum_{n=1}^{\infty}a_n\), its \(N\)th partial sum is \(S_N=\sum_{n=1}^{N}a_n\). The series converges to \(S\) when \(\lim_{N\to\infty}S_N=S\). A series is telescoping when its finite partial sums simplify through cancellation between terms from different summands.

Decomposing a Consecutive- Factor Denominator

Suppose a term has the form \(1/[n(n+1)]\), where \(n\) is a positive integer. To express it as a sum of simpler fractions, seek constants \(A\) and \(B\) such that

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

Multiplying both sides by \(n(n+1)\) gives an identity in \(n\):

$$ 1=A(n+1)+Bn=(A+B)n+A. $$

Matching the constant term and the coefficient of \(n\) gives \(A=1\) and \(A+B=0\), so \(B=-1\). Therefore, for every positive integer \(n\),

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

The opposite signs matter: the fraction with denominator \(n+1\) in one summand cancels the fraction with denominator \(n+1\) in the next summand.

From the Identity to a Telescoping Sum

Apply the decomposition to each term of a finite sum. Each summand contributes a positive fraction and a negative fraction. Writing out several terms makes the cancellation visible; after cancellation, the first positive term and the final negative term remain.

\begin{aligned} \sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+1}\right) &=\left(1-\frac12\right) +\left(\frac12-\frac13\right) +\left(\frac13-\frac14\right) +\cdots +\left(\frac1N-\frac{1}{N+1}\right)\\ &=1-\frac{1}{N+1}. \end{aligned}

This finite-sum identity holds for every positive integer \(N\). To obtain an infinite series sum, one must then take the limit of the partial sums. In this case, \(1/(N+1)\) tends to zero, so the partial sums tend to \(1\).

Key result: For every positive integer \(N\), \(\sum_{n=1}^{N}1/[n(n+1)]=1-1/(N+1)\). Consequently, \(\sum_{n=1}^{\infty}1/[n(n+1)]=1\), because the partial sums approach \(1\) as \(N\) tends to infinity.

Worked Examples

Worked Example: Sum the Series with Terms \(1/[n(n+1)]\)

Find \(\sum_{n=1}^{\infty}1/[n(n+1)]\). First decompose one term:

\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1},

since \((1/n)-(1/(n+1))=((n+1)-n)/[n(n+1)]=1/[n(n+1)]\). Thus its \(N\)th partial sum is

\begin{aligned} S_N &=\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ &=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+ \left(\frac1N-\frac{1}{N+1}\right)\\ &=1-\frac{1}{N+1}. \end{aligned}

Because \(\lim_{N\to\infty}1/(N+1)=0\), it follows that \(\lim_{N\to\infty}S_N=1\). Therefore the series converges to \(1\). As a check, the finite partial-sum formula gives \(S_1=1-1/2=1/2\), which is the original first term \(1/(1\cdot2)\), and differentiating is not relevant: this result is verified by the exact finite-sum cancellation and its limit.

Worked Example: Evaluate a Shifted Finite Sum

Evaluate \(\sum_{n=0}^{5}4/[(n+2)(n+3)]\). The two denominator factors are consecutive, so write

\frac{4}{(n+2)(n+3)} =4\left(\frac{1}{n+2}-\frac{1}{n+3}\right).

To check the decomposition, the difference inside the parentheses has numerator \((n+3)-(n+2)=1\), with denominator \((n+2)(n+3)\). Now expand the six terms:

\begin{aligned} \sum_{n=0}^{5}\frac{4}{(n+2)(n+3)} =4\bigg[ &\left(\frac12-\frac13\right) +\left(\frac13-\frac14\right) +\left(\frac14-\frac15\right)\\ &+\left(\frac15-\frac16\right) +\left(\frac16-\frac17\right) +\left(\frac17-\frac18\right) \bigg]. \end{aligned}

Every intermediate reciprocal cancels once with a positive and once with a negative sign. The remaining sum is

4\left(\frac12-\frac18\right) =4\left(\frac{4}{8}-\frac{1}{8}\right) =4\cdot\frac38 =\frac32.

Thus the finite sum is \(3/2\). The result is checked directly by the uncancelled endpoints: the first positive term is \(1/2\), the final negative term is \(-1/8\), and their difference, multiplied by \(4\), is \(3/2\).

Worked Example: A Three-Fraction Decomposition

Evaluate \(\sum_{n=1}^{\infty}2/[n(n+1)(n+2)]\). Seek constants \(A\), \(B\), and \(C\) with

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

After multiplying through by \(n(n+1)(n+2)\), the numerator identity is

$$ 2=A(n+1)(n+2)+Bn(n+2)+Cn(n+1). $$

Set \(n=0\) to obtain \(2=2A\), so \(A=1\). Set \(n=-1\) to obtain \(2=-B\), so \(B=-2\). Set \(n=-2\) to obtain \(2=2C\), so \(C=1\). Therefore,

\frac{2}{n(n+1)(n+2)} =\frac1n-\frac{2}{n+1}+\frac{1}{n+2}.

For a check, placing the right-hand side over the common denominator gives numerator \((n+1)(n+2)-2n(n+2)+n(n+1)=2\), as required. Now let \(H_m=\sum_{k=1}^{m}1/k\). Summing the decomposition from \(n=1\) to \(N\) gives

\begin{aligned} S_N &=H_N-2(H_{N+1}-1)+(H_{N+2}-1-\tfrac12)\\ &=H_N-2H_{N+1}+H_{N+2}+\frac12. \end{aligned}

Using \(H_{N+1}=H_N+1/(N+1)\) and \(H_{N+2}=H_N+1/(N+1)+1/(N+2)\), simplify:

S_N =\frac12-\frac{1}{N+1}+\frac{1}{N+2}.

Both reciprocal terms tend to zero as \(N\) tends to infinity, so \(S_N\) tends to \(1/2\). The series sum is \(1/2\). The finite partial-sum expression also checks at \(N=1\): it gives \(1/2-1/2+1/3=1/3\), equal to the original first term \(2/(1\cdot2\cdot3)=1/3\).

Why the Method Works

The essential argument can be stated for any finite sum of consecutive differences. If \(b_n\) is a sequence, then the \(n\)th term \(b_n-b_{n+1}\) has a negative part that matches the positive part of the next term. For every positive integer \(N\), distributing the sum gives

\begin{aligned} \sum_{n=1}^{N}(b_n-b_{n+1}) &=(b_1-b_2)+(b_2-b_3)+\cdots+(b_N-b_{N+1})\\ &=b_1-b_{N+1}. \end{aligned}

Each interior term \(b_2,\ldots,b_N\) occurs once positively and once negatively, so its total contribution is zero. This proves the finite telescoping identity. If \(\lim_{N\to\infty}b_{N+1}=L\), the partial sums tend to \(b_1-L\), and the infinite series converges to that value. The limit condition matters: cancellation in finite sums alone does not establish that an infinite series has a finite sum.

Factor the denominator.
Identify the linear factors, and check that the indices in the fractions you propose are defined for every term being summed.

Find and verify the coefficients.
Multiply by the common denominator, match coefficients or substitute convenient values, and check the resulting numerator identity.

Write the finite partial sum.
Expand enough terms to identify cancellations and retain all uncancelled terms at the endpoints.

Take a limit only when needed.
For an infinite series, find the limit of the partial-sum formula. A finite telescoping formula does not by itself guarantee convergence.

A Common Sign Error

A frequent mistake is to write the decomposition with two positive terms:

\frac{1}{n(n+1)} \mathrel{\ne} \frac1n+\frac{1}{n+1}.

The right-hand side has common-denominator numerator \((n+1)+n=2n+1\), not \(1\). For example, at \(n=1\), the left side is \(1/2\), while the proposed right side is \(1+1/2=3/2\). The correct sign follows by solving for the coefficients: \(1=A(n+1)+Bn\), so \(A=1\) and \(A+B=0\), which gives \(B=-1\). Always combine the proposed fractions over a common denominator to check the numerator before summing.

When This Approach Is Useful

Partial fractions are useful when a rational term has a denominator that factors into simple linear factors and the resulting fractions line up across successive values of \(n\). A successful decomposition may expose complete cancellation, partial cancellation, or no telescoping pattern at all. Therefore, do not assume that every rational series telescopes merely because its terms can be decomposed.

For an infinite series, the reliable procedure is to derive a formula for the finite partial sum first, then determine its limit. This keeps the endpoint terms visible and distinguishes a convergent series from a cancellation pattern that does not settle to a finite value.

Check Your Understanding

Use partial fractions and finite partial sums where appropriate. Check each decomposition by recombining its fractions.

Find \(A\) and \(B\) so that \(1/[(n+2)(n+3)]=A/(n+2)+B/(n+3)\).
Write the \(N\)th partial sum of \(\sum_{n=1}^{\infty}1/[n(n+1)]\), then state its limit.
Evaluate \(\sum_{n=2}^{6}3/[n(n+1)]\).
Why must the limit of the partial sums be checked after a telescoping decomposition?
What numerator results if \(1/n+1/(n+1)\) is combined over a common denominator?