Telescoping Series for MATH 141
Exam Relevance for MATH 141
Telescoping series appear occasionally in MATH 141. Recognize partial fractions that cancel.
What is a Telescoping Series?
A telescoping series is a series where most terms cancel when you write out the partial sums, leaving only a few terms at the beginning and end.
The name comes from how an old-fashioned telescope collapses — the middle sections slide into each other and disappear!
The Classic Example
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$$
Step 1: Use partial fractions
$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
Step 2: Write out the partial sum
$$S_N = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+1}\right)$$
Step 3: Watch the cancellation!
$$S_N = 1 \color{red}{- \frac{1}{2}} \color{red}{+ \frac{1}{2}} \color{blue}{- \frac{1}{3}} \color{blue}{+ \frac{1}{3}} \color{green}{- \frac{1}{4}} \color{green}{+ \frac{1}{4}} - \cdots \color{purple}{+ \frac{1}{N}} \color{purple}{- \frac{1}{N}} - \frac{1}{N+1}$$
Everything cancels except the first and last terms:
$$S_N = 1 - \frac{1}{N+1}$$
Step 4: Take the limit
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left(1 - \frac{1}{N+1}\right) = 1$$
The Pattern to Recognize
Telescoping series typically have the form:
$$\sum (b_n - b_{n+1}) \quad \text{or} \quad \sum (b_n - b_{n+k})$$
When you add these up:
- $b_1$ stays
- $b_2$ cancels with $-b_2$
- $b_3$ cancels with $-b_3$
- ...eventually only the first few and last few terms remain
Result: $S_N = b_1 - b_{N+1}$ (for the basic case)
Setting Up Telescoping with Partial Fractions
Most telescoping problems require partial fractions first.
Problem: Find the sum: $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+2)}$
Step 1: Partial fractions
$$\frac{1}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}$$
Multiply both sides by $n(n+2)$:
$$1 = A(n+2) + Bn$$
Set $n = 0$: $1 = 2A \Rightarrow A = \frac{1}{2}$
Set $n = -2$: $1 = -2B \Rightarrow B = -\frac{1}{2}$
$$\frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$$
Step 2: Write partial sum
$$S_N = \frac{1}{2}\sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+2}\right)$$
$$= \frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \cdots\right]$$
Step 3: Identify what survives
The $\frac{1}{3}$ from the first term cancels with $-\frac{1}{3}$ from the third term. The $\frac{1}{4}$ from the second term cancels with $-\frac{1}{4}$ from the fourth term. And so on.
What survives: $\frac{1}{1}$ and $\frac{1}{2}$ from the start, and $-\frac{1}{N+1}$ and $-\frac{1}{N+2}$ from the end.
$$S_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right)$$
Step 4: Take limit
$$\sum_{n=1}^{\infty} \frac{1}{n(n+2)} = \lim_{N \to \infty} \frac{1}{2}\left(\frac{3}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right) = \frac{1}{2} \cdot \frac{3}{2} = \boxed{\frac{3}{4}}$$
Problem: Find the sum: $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + \sqrt{n+1}}$
Rationalize the denominator:
$$\frac{1}{\sqrt{n} + \sqrt{n+1}} \cdot \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n}$$
Now it's obviously telescoping!
$$S_N = \sum_{n=1}^{N} \left(\sqrt{n+1} - \sqrt{n}\right)$$
$$= (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \cdots + (\sqrt{N+1} - \sqrt{N})$$
$$= \sqrt{N+1} - 1$$
$$\lim_{N \to \infty} (\sqrt{N+1} - 1) = \infty$$
$$\boxed{\text{The series diverges}}$$
Problem: Find the sum: $\displaystyle\sum_{n=1}^{\infty} \ln\left(\frac{n+1}{n}\right)$
$$\ln\left(\frac{n+1}{n}\right) = \ln(n+1) - \ln(n)$$
This telescopes!
$$S_N = \sum_{n=1}^{N} [\ln(n+1) - \ln(n)]$$
$$= [\ln 2 - \ln 1] + [\ln 3 - \ln 2] + [\ln 4 - \ln 3] + \cdots + [\ln(N+1) - \ln N]$$
$$= \ln(N+1) - \ln 1 = \ln(N+1)$$
$$\lim_{N \to \infty} \ln(N+1) = \infty$$
$$\boxed{\text{The series diverges}}$$
Problem: Find the sum: $\displaystyle\sum_{n=1}^{\infty} \arctan\left(\frac{1}{n^2+n+1}\right)$
Use the identity: $\arctan a - \arctan b = \arctan\left(\dfrac{a-b}{1+ab}\right)$
With $a = n+1$ and $b = n$:
$$\arctan(n+1) - \arctan(n) = \arctan\left(\frac{1}{1+n(n+1)}\right) = \arctan\left(\frac{1}{n^2+n+1}\right)$$
So our series telescopes!
$$S_N = \sum_{n=1}^{N} [\arctan(n+1) - \arctan(n)] = \arctan(N+1) - \arctan(1)$$
$$\lim_{N \to \infty} [\arctan(N+1) - \arctan(1)] = \frac{\pi}{2} - \frac{\pi}{4} = \boxed{\frac{\pi}{4}}$$
Problem: Find $S_5$ for $\displaystyle\sum_{n=1}^{\infty} \frac{2}{n(n+1)}$.
$$\frac{2}{n(n+1)} = 2\left(\frac{1}{n} - \frac{1}{n+1}\right)$$
$$S_N = 2\left(1 - \frac{1}{N+1}\right) = 2 \cdot \frac{N}{N+1} = \frac{2N}{N+1}$$
$$S_5 = \frac{2(5)}{5+1} = \frac{10}{6} = \boxed{\frac{5}{3}}$$
How to Recognize Telescoping
Look for:
| Form | Try |
|---|---|
| $\dfrac{1}{n(n+1)}$ | Partial fractions → $\dfrac{1}{n} - \dfrac{1}{n+1}$ |
| $\dfrac{1}{n(n+k)}$ | Partial fractions → $\dfrac{1}{k}\left(\dfrac{1}{n} - \dfrac{1}{n+k}\right)$ |
| $\dfrac{1}{\sqrt{n}+\sqrt{n+1}}$ | Rationalize |
| $\ln\left(\dfrac{n+1}{n}\right)$ | Already $\ln(n+1) - \ln(n)$ |
| Products with factorials | Check for $\dfrac{a_n}{a_{n+1}}$ patterns |
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting to track ALL surviving terms
Wrong: For $\sum \frac{1}{n(n+2)}$, saying only $\frac{1}{1}$ survives.
Why it's wrong: When the gap is 2 (not 1), TWO terms survive at the beginning and TWO at the end.
Correct: $S_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right)$. Track the gap!
❌ Mistake: Assuming all telescoping series converge
Wrong: "It telescopes, so it must converge."
Why it's wrong: Example 3 telescoped to $\sqrt{N+1} - 1$, which diverges! Example 4 telescoped to $\ln(N+1)$, also divergent.
Correct: After telescoping, you still need $\lim_{N \to \infty} S_N$ to be finite.
❌ Mistake: Skipping partial fractions
Wrong: Trying to telescope $\sum \frac{1}{n(n+1)}$ without decomposing it first.
Why it's wrong: The fraction $\frac{1}{n(n+1)}$ doesn't obviously cancel with anything. You need $\frac{1}{n} - \frac{1}{n+1}$ to see the telescoping.
Correct: Always decompose using partial fractions (or rationalize, or use log properties) to reveal the telescoping structure.
❌ Mistake: Wrong partial fraction coefficients
Wrong: $\frac{1}{n(n+2)} = \frac{1}{n} - \frac{1}{n+2}$ (missing the $\frac{1}{2}$)
Why it's wrong: Check: $\frac{1}{n} - \frac{1}{n+2} = \frac{(n+2)-n}{n(n+2)} = \frac{2}{n(n+2)} \neq \frac{1}{n(n+2)}$
Correct: $\frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$. Always verify your partial fractions!
Telescoping Series Pattern
When consecutive terms cancel, only the first and last terms survive. Take the limit as N → ∞ to find the sum (if it converges).
Variables:
- $b_n$:
- the nth term in the telescoping pattern
- $b_1$:
- first term (survives)
- $b_{N+1}$:
- last term (survives, take limit)
Common Telescoping Decomposition
The most common partial fraction setup for telescoping. The factor of 1/k is often forgotten!
Variables:
- $n$:
- the index variable
- $k$:
- the gap between terms (often 1 or 2)
Telescoping with Consecutive Integers
The classic telescoping setup. Sum equals 1 - 1/(N+1), which converges to 1.
Variables:
- $n$:
- index starting at 1
- $S_N$:
- partial sum = 1 - 1/(N+1)
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