Telescoping Series

Definition

1.A telescoping series is a series whose general term can be written as the difference of two consecutive values of an auxiliary sequence. When the terms are added, each negative part cancels the positive part of the following term, and the partial sum collapses to a few surviving terms. The name comes from a hand telescope, which folds down to a fraction of its extended length when the inner tubes slide into one another.

2.Suppose the general term of a series has the form:

$$ a_k = b_k - b_{k+1} $$

3.for some sequence $\{b_k\}$. The partial sum of the first $n$ terms then reduces to two terms:

$$ \begin{align} S_n = \sum_{k=1}^{n} a_k &= (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + \dots + (b_n - b_{n+1}) \\[6pt] &= b_1 - b_{n+1} \end{align} $$

4.Every interior term appears once with a plus sign and once with a minus sign, so only the first and the last value remain. The difficulty of a telescoping problem lies in recognizing the sequence $\{b_k\}$, since once the term is rewritten as a difference the sum is immediate.

Convergence of the infinite series

5.The behavior of the infinite series depends only on the limit of the auxiliary sequence. From the partial sum $S_n = b_1 - b_{n+1}$, the infinite series converges exactly when $\{b_{n+1}\}$ has a finite limit:

$$ \sum_{k=1}^{\infty} (b_k - b_{k+1}) = b_1 - \lim_{n \to \infty} b_{n+1} $$

6.If $b_n \to L$ as $n \to \infty$, the sum equals $b_1 - L$. If the sequence has no finite limit, the series diverges. A telescoping series therefore turns a question about an infinite sum into a question about a single limit, the limit of the sequence that generates the differences.

Example

7.We evaluate the finite sum of reciprocals of consecutive products:

$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{99 \cdot 100} $$

8.The general term is $\frac{1}{k(k+1)}$. Decomposing it into partial fractions expresses it as a difference:

$$ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} $$

9.This matches the telescoping form with $b_k = \frac{1}{k}$. Substituting the decomposition into the sum and letting the interior terms cancel gives:

$$ \begin{align} \sum_{k=1}^{99} \frac{1}{k(k+1)} &= \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \dots + \left( \frac{1}{99} - \frac{1}{100} \right) \\[6pt] &= 1 - \frac{1}{100} = \frac{99}{100} \end{align} $$

10.The same decomposition handles the infinite series. Here $b_{n+1} = \frac{1}{n+1} \to 0$, so the sum converges:

$$ \sum_{k=1}^{\infty} \frac{1}{k(k+1)} = 1 - \lim_{n \to \infty} \frac{1}{n+1} = 1 $$

11.This convergent value, the Mengoli sum, makes $\sum \frac{1}{n(n+1)}$ a standard model for the comparison test and the asymptotic comparison test, where a more involved positive-term series is matched against it.

A fixed gap between the cancelling terms

12.In the basic form each term cancels against its immediate successor, and a single value survives. The same mechanism works when the two parts of the difference are separated by a fixed number of positions. Suppose the general term has the form:

$$ a_k = b_k - b_{k+p} $$

13.for a fixed integer $p \geq 1$. The positive parts of the partial sum run through $b_1, b_2, \dots, b_n$, while the negative parts run through $b_{1+p}, b_{2+p}, \dots, b_{n+p}$. The two ranges overlap on the indices from $1+p$ to $n$, where the cancellation takes place, and a block of $p$ terms survives at each end:

$$ S_n = \sum_{k=1}^{n} (b_k - b_{k+p}) = (b_1 + b_2 + \dots + b_p) - (b_{n+1} + b_{n+2} + \dots + b_{n+p}) $$

14.The leading block $b_1 + \dots + b_p$ is fixed, so convergence depends only on the trailing block. If $b_n \to L$, each of the $p$ final terms tends to $L$, and the series sums to:

$$ \sum_{k=1}^{\infty} (b_k - b_{k+p}) = (b_1 + b_2 + \dots + b_p) - pL $$

15.When $L = 0$ the sum is the leading block $b_1 + \dots + b_p$. The consecutive case of the previous sections corresponds to $p = 1$, where the block reduces to the single term $b_1$.

16.A partial fraction decomposition often produces this pattern, with the gap fixed by the distance between the factors of the denominator. We evaluate the infinite sum:

$$ \sum_{k=1}^{\infty} \frac{1}{k(k+2)} $$

17.The factors $k$ and $k+2$ stand two positions apart, so the decomposition splits the term into a difference with gap $p = 2$:

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

18.This matches the gap form with $b_k = \frac{1}{k}$ and an overall factor $\frac{1}{2}$. Since $b_k \to 0$, only the leading block of two terms survives, and the sum is:

$$ \sum_{k=1}^{\infty} \frac{1}{k(k+2)} = \frac{1}{2} (b_1 + b_2) = \frac{1}{2} \left( 1 + \frac{1}{2} \right) = \frac{3}{4} $$

19.The factor $\frac{1}{2}$ from the decomposition multiplies the whole result, a step that disappears in the Mengoli case where the two factors are adjacent and the constant equals $1$.

A telescoping series with factorials

20.Telescoping is not restricted to fractions. Consider the sequence $a_k = (k^2 + 1) \, k!$ and the partial sums $b_k = a_1 + a_2 + \dots + a_k$. We find a closed form for $b_k$.

21.The key step rewrites the general term as a difference of consecutive multiples of factorials. Starting from $a_k$ and using $k \cdot k! = (k+1)! - k!$, we get:

$$ \begin{align} (k^2 + 1) \, k! &= k(k \cdot k!) + k! \\[6pt] &= k \left[ (k+1)! - k! \right] + k! \\[6pt] &= k(k+1)! - k \cdot k! + k! \\[6pt] &= k(k+1)! - (k-1) \, k! \end{align} $$

22.The right side is the difference $c_{k+1} - c_k$ with $c_k = (k-1) \, k!$, since $c_{k+1} = k(k+1)!$. Summing from $1$ to $k$ collapses the partial sum to its final term, because $c_1 = 0 \cdot 1! = 0$:

$$ b_k = \sum_{j=1}^{k} (c_{j+1} - c_j) = c_{k+1} - c_1 = k(k+1)! $$

23.This closed form simplifies ratios involving the partial sums. For instance, the quotient of a term and the running total is:

$$ \frac{a_k}{b_k} = \frac{(k^2 + 1) \, k!}{k(k+1)!} = \frac{k^2 + 1}{k(k+1)} = \frac{k^2 + 1}{k^2 + k} $$

24.At $k = 100$ this gives $\frac{a_{100}}{b_{100}} = \frac{10001}{10100}$, a ratio already in lowest terms.

Telescoping products

25.The cancellation that simplifies a telescoping sum has a multiplicative counterpart, where consecutive factors cancel instead of consecutive terms. We evaluate the product:

$$ \prod_{k=2}^{N} \frac{k^2}{k^2 - 1} $$

26.Factoring the denominator as $k^2 - 1 = (k-1)(k+1)$ splits each factor into two ratios:

$$ \frac{k^2}{(k-1)(k+1)} = \frac{k}{k-1} \cdot \frac{k}{k+1} $$

27.The product separates into two telescoping products, one rising and one falling:

$$ \prod_{k=2}^{N} \frac{k}{k-1} = \frac{N}{1}, \qquad \prod_{k=2}^{N} \frac{k}{k+1} = \frac{2}{N+1} $$

28.In the first product each numerator cancels the denominator of the next factor, leaving $N$ over the initial denominator $1$. In the second the cancellation runs the other way, leaving the initial numerator $2$ over the final denominator $N+1$. Multiplying the two surviving fractions gives the value of the product:

$$ \prod_{k=2}^{N} \frac{k^2}{k^2 - 1} = \frac{N}{1} \cdot \frac{2}{N+1} = \frac{2N}{N+1} $$

29.As $N \to \infty$ this approaches $2$, so the infinite product converges to $2$.