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Series and Partial Sums · Tutorial 568 of 998

Partial Fractions for Telescoping Series

Use partial fractions to expose cancellation in rational series and evaluate their finite and infinite sums.

Beginner 9 min read

What You'll Learn

  • Decompose a rational term with two linear denominator factors into simpler fractions
  • Recognize when the resulting fractions are shifted versions of one another
  • Write a finite partial sum and identify exactly which terms cancel
  • Evaluate infinite telescoping sums by taking a limit of their partial sums
  • Avoid sign errors and check a proposed decomposition by recombining its fractions

Why Partial Fractions Help with Series

A rational term can look difficult to add because its denominator is a product. Partial fractions can rewrite that term as a difference of two simpler fractions. When the fractions at consecutive indices match, most terms cancel in the partial sums. This cancellation is called telescoping.

The crucial point is that partial fractions do not, by themselves, guarantee telescoping. The simpler fractions must be shifted versions of one another. When they are, writing out a finite sum makes the cancellation visible and provides a direct route to the infinite sum, if the partial sums have a limit.

Definition: A series \(\sum_{n=m}^{\infty} a_n\) converges to \(S\) when its partial sums \(S_N=\sum_{n=m}^{N}a_n\) have limit \(S\) as \(N\) tends to infinity. A series whose partial sums cancel in a chain of matching terms is called telescoping.

Partial Fractions for Neighboring Linear Factors

Consider two linear factors with the same coefficient of \(n\), separated by one step in the index: \(an+b\) and \(a(n+1)+b\), where \(a\ne 0\). Their difference as reciprocals is especially useful:

$$ \frac{1}{an+b}-\frac{1}{a(n+1)+b} = \frac{a(n+1)+b-(an+b)} {(an+b)(a(n+1)+b)} = \frac{a}{(an+b)(a(n+1)+b)}. $$

Therefore, for a constant \(C\), the rational term with these factors can be decomposed as follows, wherever neither denominator is zero:

$$ \frac{C}{(an+b)(a(n+1)+b)} = \frac{C}{a} \left( \frac{1}{an+b}-\frac{1}{a(n+1)+b} \right). $$

This is a partial-fraction decomposition: each side has the same value, but the right-hand side displays the two simpler fractions explicitly. The coefficients have opposite signs because subtracting the fractions produces the constant difference \(a\) in the numerator.

For example, when the denominator factors are \(n\) and \(n+1\), the reciprocal difference is \(1/n-1/(n+1)\). When the factors are \(3n+2\) and \(3n+5\), they are also neighbors in this sense: the second factor is \(3(n+1)+2\). The shared coefficient of \(n\) and the one-step shift are what make the cancellation work.

How the Cancellation Produces a Sum

Suppose a term has the displayed form and the index runs from \(m\) through \(N\). Substituting the decomposition into the finite partial sum gives one positive reciprocal and one negative reciprocal at each index. Expanding a few terms shows the pattern:

$$ \begin{aligned} \sum_{n=m}^{N}\frac{C}{(an+b)(a(n+1)+b)} &=\frac{C}{a}\sum_{n=m}^{N} \left(\frac{1}{an+b}-\frac{1}{a(n+1)+b}\right)\\ &=\frac{C}{a}\left[ \left(\frac{1}{am+b}-\frac{1}{a(m+1)+b}\right) +\left(\frac{1}{a(m+1)+b}-\frac{1}{a(m+2)+b}\right) +\cdots\right.\\ &\qquad\left.+ \left(\frac{1}{aN+b}-\frac{1}{a(N+1)+b}\right) \right]. \end{aligned} $$

Every negative term except the last is canceled by the identical positive term in the next summand. The first positive term and the final negative term remain.

$$ \sum_{n=m}^{N}\frac{C}{(an+b)(a(n+1)+b)} = \frac{C}{a} \left( \frac{1}{am+b}-\frac{1}{a(N+1)+b} \right). $$

This identity is a finite-sum formula. It holds when the denominators in the terms are nonzero for the indices being summed. If the final reciprocal tends to zero as \(N\) tends to infinity, the corresponding infinite series converges to the limit of these partial sums.

Key takeaway: The finite partial sum has two endpoint terms: one from the first positive fraction and one from the final negative fraction. For an infinite sum, take the limit of the partial-sum formula; do not simply erase the uncanceled endpoint without checking its limit.

Worked Example: The Basic Reciprocal Product

Worked Example: The Basic Reciprocal Product

Evaluate \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\) by partial fractions.

First find constants \(A\) and \(B\) such that

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

Putting the right-hand side over the common denominator \(n(n+1)\) gives numerator \(A(n+1)+Bn=(A+B)n+A\). For this to equal the constant numerator \(1\), the coefficient equations are \(A+B=0\) and \(A=1\). Thus \(A=1\) and \(B=-1\), so

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

Check the decomposition by recombining the fractions:

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

For a finite upper index \(N\), expand the partial sum:

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

As \(N\) tends to infinity, \(1/(N+1)\) tends to zero. Therefore the series converges, and its sum is

$$ \sum_{n=1}^{\infty}\frac{1}{n(n+1)} = \lim_{N\to\infty}\left(1-\frac{1}{N+1}\right) =1. $$

The result checks against the partial-sum formula: each finite sum is \(1-1/(N+1)\), which is less than \(1\) and approaches \(1\).

Worked Example: A Finite Sum with Scaled Factors

Worked Example: A Finite Sum with Scaled Factors

Find \(\displaystyle\sum_{n=1}^{N}\frac{7}{(3n+2)(3n+5)}\).

The second factor is \(3(n+1)+2\), so the reciprocal difference is

$$ \frac{1}{3n+2}-\frac{1}{3n+5} = \frac{(3n+5)-(3n+2)}{(3n+2)(3n+5)} = \frac{3}{(3n+2)(3n+5)}. $$

Multiplying this identity by \(7/3\) gives the required term:

$$ \frac{7}{(3n+2)(3n+5)} = \frac{7}{3} \left( \frac{1}{3n+2}-\frac{1}{3n+5} \right). $$

Recombining verifies the decomposition: the numerator on the right is \((7/3)((3n+5)-(3n+2))=(7/3)(3)=7\), over the original denominator.

Now sum the difference from \(n=1\) to \(N\). The negative term \(1/(3n+5)\) at index \(n\) equals the positive term \(1/(3(n+1)+2)\) at index \(n+1\). Only the first and last terms remain:

$$ \begin{aligned} \sum_{n=1}^{N}\frac{7}{(3n+2)(3n+5)} &=\frac{7}{3} \left( \frac15-\frac{1}{3N+5} \right). \end{aligned} $$

As a numerical check, take \(N=2\). The original two terms add to

$$ \frac{7}{5\cdot 8}+\frac{7}{8\cdot 11} = \frac{7}{40}+\frac{7}{88} = \frac{77}{440}+\frac{35}{440} = \frac{112}{440} = \frac{14}{55}. $$

The formula gives the same result:

$$ \frac{7}{3}\left(\frac15-\frac1{11}\right) = \frac{7}{3}\left(\frac{6}{55}\right) = \frac{14}{55}. $$

Worked Example: An Infinite Sum with Odd-Spaced Denominators

Worked Example: An Infinite Sum with Odd-Spaced Denominators

Evaluate \(\displaystyle\sum_{n=1}^{\infty}\frac{8}{(4n-1)(4n+3)}\).

The factors are \(4n-1\) and \(4(n+1)-1=4n+3\). Their reciprocal difference is

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

Consequently, the original term is twice this difference:

$$ \frac{8}{(4n-1)(4n+3)} = 2\left(\frac{1}{4n-1}-\frac{1}{4n+3}\right). $$

Recombining gives numerator \(2((4n+3)-(4n-1))=2\cdot4=8\), confirming the decomposition. The finite partial sum is

$$ \begin{aligned} \sum_{n=1}^{N}\frac{8}{(4n-1)(4n+3)} &=2\left(\frac13-\frac{1}{4N+3}\right). \end{aligned} $$

For a direct check at \(N=2\), the original terms give

$$ \frac{8}{3\cdot7}+\frac{8}{7\cdot11} = \frac{8}{21}+\frac{8}{77} = \frac{88}{231}+\frac{24}{231} = \frac{112}{231} = \frac{16}{33}. $$

The finite-sum formula gives \(2(1/3-1/11)=2(8/33)=16/33\), as required. Finally, \(1/(4N+3)\) tends to zero, so

$$ \sum_{n=1}^{\infty}\frac{8}{(4n-1)(4n+3)} = \lim_{N\to\infty}2\left(\frac13-\frac{1}{4N+3}\right) = \frac23. $$

Here the infinite sum follows from the limit of the finite partial sums, not from cancellation alone. The remaining endpoint is explicitly shown to approach zero.

A Common Sign Error

A frequent mistake is to use a plus sign where the partial fractions require a minus sign. For instance, the incorrect calculation

$$ \frac{1}{n(n+1)} \overset{\text{wrong}}{=} \frac1n+\frac{1}{n+1} $$

does not reproduce the original numerator. Combining the fractions on the right gives

$$ \frac1n+\frac{1}{n+1} = \frac{(n+1)+n}{n(n+1)} = \frac{2n+1}{n(n+1)}, $$

which is not \(1/[n(n+1)]\). The correction is to subtract:

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

Check the numerator whenever you find partial fractions. A sign error changes the terms being summed, and the apparent cancellation cannot fix an incorrect decomposition.

When This Method Applies

The method is effective when the denominator factors are linear and the resulting fractions at consecutive indices match. In particular, a product of the form \((an+b)(a(n+1)+b)\) gives exactly the reciprocal difference used above. The factors must not vanish at any index in the sum.

After writing a proposed decomposition, verify it by combining the fractions over their common denominator. Then write out the finite partial sum, identify the uncanceled endpoints, and only then take a limit if the series is infinite. If the endpoint does not tend to a finite value, this calculation does not establish convergence to the claimed sum.

Partial fractions are useful here because they transform a product in the denominator into a difference whose cancellation is visible. The finite-sum identity records precisely what survives, while the limit of those partial sums determines the infinite sum when that limit exists.

Check Your Understanding

Use the decomposition and finite-sum reasoning in each question. For infinite sums, state the endpoint limit you use.

  1. Find a partial-fraction decomposition of \(1/[(n+2)(n+3)]\), then write its finite sum from \(n=1\) to \(N\).
  2. Evaluate \(\displaystyle\sum_{n=1}^{\infty}2/[(n+1)(n+2)]\).
  3. Rewrite \(5/[(2n-1)(2n+1)]\) as a difference of reciprocals and identify its coefficient.
  4. For the sum in question 3 from \(n=1\) to \(N\), which terms remain after cancellation?
  5. Why should a proposed partial-fraction decomposition be recombined over a common denominator before using it in a series?

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