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Basic Series Divergence · Tutorial 593 of 998

Telescoping Series Workshop With Finite Remainders

Learn to track the uncancelled boundary terms, calculate exact remainders, and decide whether a telescoping series converges.

Beginner 9 min read

What You'll Learn

  • Write telescoping terms as differences of consecutive boundary terms
  • Derive finite partial-sum formulas without losing endpoint terms
  • Compute exact finite tails and remainders for convergent series
  • Evaluate a rational telescoping series using partial fractions
  • Simplify factorial terms into consecutive differences
  • Detect divergence when the boundary terms fail to settle

Why the Finite Sum Matters

A telescoping series can look complicated term by term, but many of its terms cancel when added. The key is to calculate a finite partial sum first. That calculation reveals exactly which terms survive at the boundaries; only then can we determine what happens as the number of terms increases.

In this workshop, we will use exact finite formulas to study rational and factorial examples. Those formulas also tell us how much of a convergent sum remains after its first \(N\) terms, and they expose cases where the boundary terms do not settle and the series therefore diverges.

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 \(S\) if the sequence of partial sums \(S_N\) tends to the finite limit \(S\) as \(N\to\infty\).

The Finite Cancellation Formula

Suppose each term can be written as a difference \(a_n=c_n-c_{n+1}\), where \(c_n\) is a sequence. Adding the first \(N\) terms gives

$$ S_N=(c_1-c_2)+(c_2-c_3)+\cdots+(c_N-c_{N+1})=c_1-c_{N+1}. $$

Every interior term appears once with a plus sign and once with a minus sign. The only terms left are the first positive boundary term and the final negative boundary term. This identity is finite and exact: it does not assume that the infinite series converges.

If \(c_n\) tends to a finite limit \(L\), then the partial sums tend to \(c_1-L\). Conversely, if the partial sums converge, then \(c_{N+1}=c_1-S_N\) also converges. Thus, for a series of consecutive differences, convergence is equivalent to the boundary sequence \(c_n\) having a finite limit.

Theorem: If \(a_n=c_n-c_{n+1}\) for every \(n\geq 1\), then \(S_N=c_1-c_{N+1}\). The infinite series converges precisely when \(c_n\) tends to a finite limit \(L\); in that case its sum is \(c_1-L\).

For a convergent series, the remainder after the first \(N\) terms is \(R_N=S-S_N\). Substituting the formulas for \(S\) and \(S_N\) gives \(R_N=c_{N+1}-L\). A related finite tail, from term \(N+1\) through term \(M\), is

$$ \sum_{n=N+1}^{M}a_n=c_{N+1}-c_{M+1}. $$

The finite-tail identity holds whether or not the infinite series converges. The infinite remainder formula requires convergence, because it uses the limit \(L\).

Worked Example: A Rational Series With a Two-Term Shift

Worked Example: A Rational Series With a Two-Term Shift

Find the partial sum, sum, and remainder after \(N\) terms for \(\sum_{n=1}^{\infty}\frac{1}{n(n+2)}\).

First split the rational term into consecutive boundary contributions separated by two indices. Since \(n+2-n=2\),

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

Therefore, the first \(N\) terms give

$$ \begin{aligned} S_N &=\frac12\left[ \left(1-\frac13\right) +\left(\frac12-\frac14\right) +\left(\frac13-\frac15\right) +\cdots +\left(\frac1N-\frac1{N+2}\right) \right]\\ &=\frac12\left(1+\frac12-\frac1{N+1}-\frac1{N+2}\right)\\ &=\frac34-\frac{1}{2(N+1)}-\frac{1}{2(N+2)}. \end{aligned} $$

The terms \(1/3\) through \(1/N\) cancel against the matching negative terms. The two initial terms \(1\) and \(1/2\) remain, as do the two final negative terms. As \(N\to\infty\), both fractions involving \(N\) tend to zero, so the series converges to \(3/4\). Its remainder is

$$ R_N=\frac34-S_N=\frac{1}{2(N+1)}+\frac{1}{2(N+2)}. $$

For a check with \(N=3\), the direct partial sum is \(1/3+1/8+1/15\). Using denominator \(120\), this is \(40/120+15/120+8/120=63/120=21/40\). The formula gives \(3/4-1/8-1/10=30/40-5/40-4/40=21/40\), as required. The remainder formula at \(N=3\) gives \(1/8+1/10=9/40\), and \(21/40+9/40=3/4\), verifying the total.

Worked Example: A Factorial Series

Worked Example: A Factorial Series

Find the sum and the remainder after \(N\) terms for \(\sum_{n=1}^{\infty}\frac{n}{(n+1)!}\).

Using \((n+1)!= (n+1)n!\), rewrite the numerator \(n\) as \((n+1)-1\):

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

Thus each term is the difference of consecutive factorial reciprocals, and

$$ \begin{aligned} S_N &=\left(1-\frac1{2!}\right) +\left(\frac1{2!}-\frac1{3!}\right) +\cdots +\left(\frac1{N!}-\frac1{(N+1)!}\right)\\ &=1-\frac1{(N+1)!}. \end{aligned} $$

Since \(1/(N+1)!\to 0\), the series converges to \(1\). Subtracting the partial sum from the total gives the exact remainder \(R_N=1/(N+1)!\).

For \(N=4\), the terms are \(1/2\), \(2/6=1/3\), \(3/24=1/8\), and \(4/120=1/30\). Their sum is \(15/30+10/30+3.75/30+1/30\), which is not a convenient integer-denominator calculation because \(3.75/30\) is not an integer fraction. Instead, using denominator \(120\), the terms are \(60/120+40/120+15/120+4/120=119/120\). The finite formula gives \(1-1/5!=1-1/120=119/120\). The remainder is \(1/120\), and \(119/120+1/120=1\), which checks both the partial sum and the remainder.

When the Boundary Terms Do Not Settle

The cancellation identity does not guarantee convergence. It reduces the question to the behavior of the boundary term \(c_{N+1}\). If that term has no finite limit, then \(S_N=c_1-c_{N+1}\) cannot have a finite limit either.

Worked Example: Factorial Boundaries That Oscillate

Consider the series with terms \(a_n=c_n-c_{n+1}\), where

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

Its first boundary values are \(c_1=-1/2\), \(c_2=2/3\), and \(c_3=-6/7\). For example, the first term of the series is \(a_1=c_1-c_2=-1/2-2/3=-7/6\), and the second is \(a_2=c_2-c_3=2/3-(-6/7)=32/21\). The first two partial sums therefore give

$$ S_1=-\frac76,\qquad S_2=-\frac76+\frac{32}{21}=\frac5{14}. $$

For every \(N\), however, the full finite sum is exactly \(S_N=c_1-c_{N+1}\). The factor \(n!/(n!+1)\) tends to \(1\), since its difference from \(1\) is \(1/(n!+1)\), which tends to zero. Along even indices, \(c_n\) therefore tends to \(1\); along odd indices, it tends to \(-1\). Hence \(c_n\) has no limit.

In particular, for odd \(N\), \(N+1\) is even, so the corresponding partial sums tend to \(-1/2-1=-3/2\). For even \(N\), \(N+1\) is odd, so those partial sums tend to \(-1/2-(-1)=1/2\). The two subsequences have different limits, and the series diverges. The check comes directly from the exact partial-sum formula: the nonsettling boundary values force the partial sums not to settle.

A Common Cancellation Error

When cancellation is done mentally, it is easy to remove too many terms at one end. For the rational example above, an incorrect calculation is

$$ \sum_{n=1}^{N}\frac{1}{n(n+2)} \overset{\text{wrong}}{=} \frac12\left(1-\frac1{N+2}\right). $$

This drops the uncancelled \(1/2\) at the beginning and the uncancelled \(-1/(N+1)\) at the end. For \(N=3\), it predicts \(\frac12(1-1/5)=2/5\), while the actual partial sum is \(21/40\). The correct endpoint accounting is

$$ S_N=\frac12\left(1+\frac12-\frac1{N+1}-\frac1{N+2}\right). $$

A reliable check is to expand several terms of the finite sum and mark which terms cancel. Do not infer an infinite sum until the finite boundary formula is correct.

Using Finite Remainders Effectively

Exact remainders make a convergence result more informative. In the rational example, the remainder decreases according to an explicit formula involving \(N+1\) and \(N+2\). In the factorial example, it is exactly \(1/(N+1)!\). These are not estimates: they are equalities for every \(N\). They describe precisely the difference between the infinite sum and its \(N\)-term approximation.

For a general telescoping series \(a_n=c_n-c_{n+1}\), first compute the finite sum and identify both surviving boundaries. If \(c_n\to L\), the sum is \(c_1-L\) and the remainder after \(N\) terms is \(c_{N+1}-L\). If \(c_n\) does not have a finite limit, the partial sums do not converge. This procedure handles cancellation and convergence as separate steps, so a series is never declared convergent merely because its terms look as though they cancel.

Check Your Understanding

Use finite partial sums and boundary terms to answer these questions.

  1. Write \(\frac{1}{(n+1)(n+2)}\) as a difference of consecutive reciprocal terms, then find the first \(N\) terms as a boundary expression.
  2. If \(a_n=c_n-c_{n+1}\) and \(c_n\to L\), what are the sum of the series and the remainder after \(N\) terms?
  3. For the factorial series in the worked example, calculate \(S_3\) and \(R_3\), then check that their sum is \(1\).
  4. Why does a sequence of boundary terms with two different subsequential limits prevent the associated telescoping series from converging?
  5. For a finite tail from term \(N+1\) through term \(M\), state the exact boundary formula.

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