Interactive Demo: Visualize Telescoping Series
See how terms cancel in a telescoping series using the method of differences.
Real-Life Hook: The Collapsing Tower
Imagine building a tower where each block cancels out part of the previous one. You have blocks labeled:
- Block 1: +10 -5
- Block 2: +5 -2
- Block 3: +2 -1
- Block 4: +1 -0.5
When you stack them, the -5 from Block 1 cancels with +5 from Block 2, the -2 from Block 2 cancels with +2 from Block 3, and so on!
Final result: Only +10 from first block and -0.5 from last block remain = 9.5
This is the Method of Differences - a powerful telescoping technique that makes seemingly impossible summations trivial!
What is the Method of Differences?
The Method of Differences (also called Telescoping Series) is used when the general term of a series can be expressed as:
$$T_n = f(n+1) - f(n)$$When we sum such terms, most terms cancel out, leaving only the first and last terms!
$$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} [f(k+1) - f(k)]$$Expanding:
$$S_n = [f(2) - f(1)] + [f(3) - f(2)] + [f(4) - f(3)] + \cdots + [f(n+1) - f(n)]$$After cancellation:
$$\boxed{S_n = f(n+1) - f(1)}$$Key Insight: Find function $f$ such that $T_n = f(n+1) - f(n)$!
Core Technique: Finding the Difference Function
Strategy: Use Partial Fractions
Many series can be expressed as differences using partial fractions.
Example 1: $T_n = \frac{1}{n(n+1)}$
Using partial fractions:
$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$So $f(n) = \frac{1}{n}$ and:
$$S_n = \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right) = \frac{1}{1} - \frac{1}{n+1} = \frac{n}{n+1}$$Example 2: $T_n = \frac{1}{n(n+1)(n+2)}$
Using partial fractions:
$$\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right]$$So $f(n) = \frac{1}{2n(n+1)}$ and:
$$S_n = \frac{1}{2}\left[\frac{1}{1 \times 2} - \frac{1}{(n+1)(n+2)}\right] = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}$$Standard Forms
Form 1: $\sum \frac{1}{n(n+1)}$
$$\boxed{\sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{n}{n+1}}$$Partial Fraction: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$
Form 2: $\sum \frac{1}{n(n+2)}$
$$\boxed{\sum_{k=1}^{n} \frac{1}{k(k+2)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}}$$Partial Fraction: $\frac{1}{k(k+2)} = \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right)$
Telescoping (with gap of 2):
$$S_n = \frac{1}{2}\left[\left(\frac{1}{1} + \frac{1}{2}\right) - \left(\frac{1}{n+1} + \frac{1}{n+2}\right)\right]$$Form 3: $\sum \frac{1}{n(n+1)(n+2)}$
$$\boxed{\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)} = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}}$$Partial Fraction: $\frac{1}{k(k+1)(k+2)} = \frac{1}{2}\left[\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}\right]$
Form 4: $\sum \frac{1}{\sqrt{n} + \sqrt{n+1}}$
$$\boxed{\sum_{k=1}^{n} \frac{1}{\sqrt{k} + \sqrt{k+1}} = \sqrt{n+1} - 1}$$Rationalization:
$$\frac{1}{\sqrt{k} + \sqrt{k+1}} = \frac{\sqrt{k+1} - \sqrt{k}}{(\sqrt{k+1} - \sqrt{k})(\sqrt{k+1} + \sqrt{k})} = \sqrt{k+1} - \sqrt{k}$$Form 5: $\sum n \cdot n!$
$$\boxed{\sum_{k=1}^{n} k \cdot k! = (n+1)! - 1}$$Recognition: $k \cdot k! = (k+1)! - k! = (k+1-1) \cdot k! = (k+1)! - k!$
The V_n Method
The V_n method is a variant used when:
$$T_n = V_{n+1} - V_n$$where $V_n$ is some function of $n$.
Process:
- Identify or construct $V_n$ such that $T_n = V_{n+1} - V_n$
- Sum: $S_n = V_{n+1} - V_1$
When to Use: When the general term involves factorials, products, or complex recursions.
Common Mistakes & How to Avoid Them
❌ Mistake 1: Incorrect Cancellation
Wrong: Assuming all terms cancel completely ✗
Correct: First few and last few terms remain! ✓
Example: $\sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+2}\right)$
Wrong: Everything cancels to 0 ✗
Correct:
$$S_n = \left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots$$Only $\frac{1}{1}$ and $\frac{1}{2}$ don’t cancel from the start, and $\frac{1}{n+1}, \frac{1}{n+2}$ don’t cancel from the end!
$$S_n = \frac{1}{1} + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} = \frac{3}{2} - \frac{2n+3}{(n+1)(n+2)}$$❌ Mistake 2: Partial Fraction Errors
Wrong: $\frac{1}{n(n+1)} = \frac{1}{n} + \frac{1}{n+1}$ ✗
Correct: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ ✓ (subtraction, not addition!)
Verification: $\frac{1}{n} - \frac{1}{n+1} = \frac{n+1-n}{n(n+1)} = \frac{1}{n(n+1)}$ ✓
❌ Mistake 3: Forgetting Initial/Final Terms
Problem: Find $\sum_{k=1}^{100} (k+1)! - k!$
Wrong: $S = 101! - 1!$ ✗ (using $f(101) - f(1)$ incorrectly)
Correct:
$$T_k = (k+1)! - k!$$Here already in difference form with $f(k) = k!$
$$S = \sum [f(k+1) - f(k)] = f(101) - f(1) = 101! - 1!$$Actually that’s correct! So no mistake here. Let me find a real mistake:
Wrong: $S = 101!$ ✗ (forgetting to subtract $1!$)
Correct: $S = 101! - 1 = 101! - 1$ ✓
Problem-Solving Strategies
Strategy 1: Recognize Partial Fractions
Problem: Find $\sum_{k=1}^{n} \frac{2}{(2k-1)(2k+1)}$
Solution:
Partial fractions:
$$\frac{2}{(2k-1)(2k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1}$$ $$2 = A(2k+1) + B(2k-1)$$Setting $k = \frac{1}{2}$: $2 = 2A$ → $A = 1$
Setting $k = -\frac{1}{2}$: $2 = -2B$ → $B = -1$
$$\frac{2}{(2k-1)(2k+1)} = \frac{1}{2k-1} - \frac{1}{2k+1}$$Telescoping:
$$S_n = \sum_{k=1}^{n} \left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)$$ $$= \left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \cdots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)$$ $$= 1 - \frac{1}{2n+1} = \frac{2n}{2n+1}$$Answer: $\frac{2n}{2n+1}$
Strategy 2: Rationalization
Problem: Find $\sum_{k=1}^{50} \frac{1}{\sqrt{k+1} + \sqrt{k}}$
Solution:
Rationalize:
$$\frac{1}{\sqrt{k+1} + \sqrt{k}} \times \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} = \frac{\sqrt{k+1} - \sqrt{k}}{(k+1) - k} = \sqrt{k+1} - \sqrt{k}$$Telescoping:
$$S_{50} = \sum_{k=1}^{50} (\sqrt{k+1} - \sqrt{k})$$ $$= (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{51} - \sqrt{50})$$ $$= \sqrt{51} - 1$$Answer: $\sqrt{51} - 1$
Strategy 3: Factorial Recognition
Problem: Find $\sum_{k=1}^{n} k \cdot k!$
Solution:
Recognize: $k \cdot k! = (k+1-1) \cdot k! = (k+1) \cdot k! - k! = (k+1)! - k!$
Telescoping:
$$S_n = \sum_{k=1}^{n} [(k+1)! - k!]$$ $$= [2! - 1!] + [3! - 2!] + [4! - 3!] + \cdots + [(n+1)! - n!]$$ $$= (n+1)! - 1!$$ $$= (n+1)! - 1$$Answer: $(n+1)! - 1$
Strategy 4: Three-Term Partial Fractions
Problem: Find $\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}$
Solution:
Partial fractions (method: decompose into two two-term fractions):
$$\frac{1}{k(k+1)(k+2)} = \frac{1}{2} \cdot \frac{2}{k(k+1)(k+2)}$$ $$= \frac{1}{2} \cdot \frac{(k+2) - k}{k(k+1)(k+2)}$$ $$= \frac{1}{2}\left[\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}\right]$$Telescoping:
$$S_n = \frac{1}{2}\left[\frac{1}{1 \times 2} - \frac{1}{(n+1)(n+2)}\right]$$ $$= \frac{1}{2}\left[\frac{1}{2} - \frac{1}{(n+1)(n+2)}\right]$$ $$= \frac{1}{4} - \frac{1}{2(n+1)(n+2)}$$Answer: $\frac{1}{4} - \frac{1}{2(n+1)(n+2)}$
Practice Problems
Level 1: JEE Main Basics
Problem 1: Find $\sum_{k=1}^{10} \frac{1}{k(k+1)}$
Solution
$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$
$S_{10} = \sum_{k=1}^{10} \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{11} = \frac{10}{11}$
Answer: $\frac{10}{11}$
Problem 2: Find $\sum_{k=1}^{n} (\sqrt{k+1} - \sqrt{k})$
Solution
This is already in telescoping form:
$S_n = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{n+1} - \sqrt{n})$
$= \sqrt{n+1} - 1$
Answer: $\sqrt{n+1} - 1$
Problem 3: Find $\sum_{k=1}^{5} \frac{1}{\sqrt{k} + \sqrt{k+1}}$
Solution
Rationalize: $\frac{1}{\sqrt{k} + \sqrt{k+1}} = \sqrt{k+1} - \sqrt{k}$
$S_5 = \sum_{k=1}^{5} (\sqrt{k+1} - \sqrt{k}) = \sqrt{6} - 1$
Answer: $\sqrt{6} - 1$
Level 2: JEE Main/Advanced
Problem 4: Find $\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}$
Solution
Partial fractions:
$$\frac{1}{(2k-1)(2k+1)} = \frac{1}{2}\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)$$$S_n = \frac{1}{2}\sum_{k=1}^{n} \left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)$
$= \frac{1}{2}\left(1 - \frac{1}{2n+1}\right)$
$= \frac{1}{2} \cdot \frac{2n}{2n+1}$
$= \frac{n}{2n+1}$
Answer: $\frac{n}{2n+1}$
Problem 5: Find $\sum_{k=2}^{n} \frac{1}{k^2-1}$
Solution
Note: $k^2 - 1 = (k-1)(k+1)$
Partial fractions:
$$\frac{1}{(k-1)(k+1)} = \frac{1}{2}\left(\frac{1}{k-1} - \frac{1}{k+1}\right)$$$S = \frac{1}{2}\sum_{k=2}^{n} \left(\frac{1}{k-1} - \frac{1}{k+1}\right)$
This telescopes with a gap of 2:
$= \frac{1}{2}\left[\left(\frac{1}{1} + \frac{1}{2}\right) - \left(\frac{1}{n} + \frac{1}{n+1}\right)\right]$
$= \frac{1}{2}\left[\frac{3}{2} - \frac{2n+1}{n(n+1)}\right]$
$= \frac{3}{4} - \frac{2n+1}{2n(n+1)}$
Answer: $\frac{3}{4} - \frac{2n+1}{2n(n+1)}$
Problem 6: Find $\sum_{k=1}^{n} k(k+1)$
Solution
Method 1: Expand and use standard formulas
$k(k+1) = k^2 + k$
$\sum k(k+1) = \sum k^2 + \sum k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$
$= \frac{n(n+1)}{6}[2n+1+3] = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}$
Method 2: Recognize as difference
$k(k+1) = \frac{(k+2)(k+1)k - (k+1)k(k-1)}{3}$… (complex!)
Method 1 is simpler for this case.
Answer: $\frac{n(n+1)(n+2)}{3}$
Level 3: JEE Advanced
Problem 7: Find $\sum_{k=1}^{n} \frac{k}{(k+1)!}$
Solution
Recognize: $\frac{k}{(k+1)!} = \frac{(k+1)-1}{(k+1)!} = \frac{1}{k!} - \frac{1}{(k+1)!}$
Telescoping: $S_n = \sum_{k=1}^{n} \left(\frac{1}{k!} - \frac{1}{(k+1)!}\right)$
$= \frac{1}{1!} - \frac{1}{(n+1)!}$
$= 1 - \frac{1}{(n+1)!}$
Answer: $1 - \frac{1}{(n+1)!}$
Problem 8: Find $\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)(k+3)}$
Solution
Partial fractions (using the trick of writing as difference of products):
$$\frac{1}{k(k+1)(k+2)(k+3)} = \frac{1}{3}\left[\frac{1}{k(k+1)(k+2)} - \frac{1}{(k+1)(k+2)(k+3)}\right]$$This telescopes!
$S_n = \frac{1}{3}\left[\frac{1}{1 \times 2 \times 3} - \frac{1}{(n+1)(n+2)(n+3)}\right]$
$= \frac{1}{3}\left[\frac{1}{6} - \frac{1}{(n+1)(n+2)(n+3)}\right]$
$= \frac{1}{18} - \frac{1}{3(n+1)(n+2)(n+3)}$
Answer: $\frac{1}{18} - \frac{1}{3(n+1)(n+2)(n+3)}$
Advanced Techniques
Technique 1: Creating Differences
Sometimes you need to create a difference by adding and subtracting terms.
Example: Sum $\sum k^2$
Recognize: $(k+1)^3 - k^3 = 3k^2 + 3k + 1$
Rearrange: $k^2 = \frac{1}{3}[(k+1)^3 - k^3 - 3k - 1]$
Then telescoping gives the standard formula.
Technique 2: Double Telescoping
Some series telescope in multiple ways simultaneously.
Example: $\sum \frac{1}{k^2(k+1)^2}$
Can be written as: $\left(\frac{1}{k} - \frac{1}{k+1}\right)^2 + \frac{1}{k(k+1)} - \frac{1}{(k)(k+1)}$… (complex!)
Cross-Topic Connections
1. Link to Partial Fractions
Essential algebra technique from rational expressions. See [Algebra - Partial Fractions].
2. Link to Limits
As $n \to \infty$, many telescoping series converge:
$$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k(k+1)} = \lim_{n \to \infty} \frac{n}{n+1} = 1$$See Limits.
3. Link to Integration
Telescoping sums are discrete analogs of the Fundamental Theorem of Calculus:
$$\sum_{k=a}^{b} [f(k+1) - f(k)] = f(b+1) - f(a)$$ $$\int_a^b f'(x) \, dx = f(b) - f(a)$$Quick Revision Checklist
- Recognize telescoping: $T_n = f(n+1) - f(n)$
- Sum = $f(n+1) - f(1)$ (last minus first!)
- Partial fractions: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$
- Rationalization: $\frac{1}{\sqrt{k+1} + \sqrt{k}} = \sqrt{k+1} - \sqrt{k}$
- Factorial form: $k \cdot k! = (k+1)! - k!$
- Watch for gap cancellation (e.g., $\frac{1}{k} - \frac{1}{k+2}$)
- Verify with first few terms to catch errors
Memory Palace Technique
Imagine a domino chain (telescoping):
- Setup (Recognize): Arrange dominoes as differences $f(k+1) - f(k)$
- Knock First (Start sum): First domino falls: $f(2) - f(1)$
- Cascade (Middle cancels): Middle dominoes knock each other: $-f(2) + f(2)$ cancels!
- Last Standing (Final result): Only first piece of first domino and last piece of last domino remain: $f(n+1) - f(1)$
Final Tips for JEE
- Master partial fractions - 90% of telescoping problems use this!
- Look for $(k+1) - k$ patterns - sign of telescoping
- Write out first 3-4 terms - visualize the cancellation
- Don’t forget $- f(1)$ - common error to only write $f(n+1)$
- Rationalization for surds - multiply by conjugate
- Factorial manipulation - $k \cdot k! = (k+1)! - k!$
- Gap telescoping - some series cancel with gap of 2 or more
- Verify answer - substitute small $n$ to check
Last Updated: October 27, 2025