How To Find Sum Of Telescoping Series

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If you’re looking at a complex-looking series and wondering how to find sum of telescoping series, you’re in the right place. This technique is a powerful tool that simplifies what seems impossible into something very manageable.

Telescoping series are special. They have a unique structure where most terms cancel out, leaving only a few to sum. This makes finding their sum much easier than adding up hundreds of terms. Once you recognize the pattern, the calculation becomes straightforward.

This guide will walk you through the process step-by-step. We’ll cover identification, the partial fraction decomposition method, and plenty of examples. You’ll learn to spot these series and sum them with confidence.

How To Find Sum Of Telescoping Series

At its core, a telescoping series is one where consecutive terms cancel each other out. The name comes from how a collapsible telescope folds in on itself. Similarly, the series “collapses” to just a few terms from the beginning and end.

The general strategy has three main steps. First, you need to express the series term in a special form, often using partial fractions. Second, you write out several terms to see the cancellation pattern. Third, you take the limit as the number of terms goes to infinity to find the sum.

What Makes a Series “Telescope”?

A series telescopes if its nth term, an, can be written as a difference: an = bn – bn+1 or something similar. When you sum from n=1 to N, the sum becomes SN = b1 – bN+1.

Notice the massive cancellation. All the intermediate terms (b2, b3, …, bN) disappear. This leaves only the first and last piece of the “telescope.” For an infinite series, you then examine the limit of bN+1 as N approaches infinity.

The Classic Example

Consider the series: ∑n=1∞ 1/(n(n+1)).

Using partial fractions: 1/(n(n+1)) = 1/n – 1/(n+1).

Let’s write the partial sum SN:

  • n=1: 1/1 – 1/2
  • n=2: 1/2 – 1/3
  • n=3: 1/3 – 1/4
  • …
  • n=N: 1/N – 1/(N+1)

Adding vertically, every term from -1/2 to 1/N cancels. So SN = 1 – 1/(N+1). The infinite sum is the limit: limN→∞ (1 – 1/(N+1)) = 1.

Step-by-Step Method to Find the Sum

Follow these numbered steps every time you suspect a series might telescope.

  1. Identify the Form. Look for series where the term is a rational function (a fraction of polynomials). Terms with products in the denominator like n(n+1), n(n+2), (2n-1)(2n+1) are prime candidates.
  2. Apply Partial Fraction Decomposition. This is the most crucial algebraic step. You break the complex fraction into a sum or difference of simpler fractions. The goal is to get it into the form bn – bn+k, where k is a positive integer (often 1).
  3. Write Out Partial Sum SN. Don’t skip this. Explicitly write the terms for n=1, 2, 3, …, N using your decomposed form. This makes the cancellation pattern visually clear and helps avoid errors.
  4. Observe the Cancellation. Add the terms vertically. Most interior terms will cancel. What remains is usually the first few terms of bn and the last few terms of bn.
  5. Simplify the Remaining Expression. After cancellation, you’ll have a formula for the Nth partial sum, SN, in terms of N. Simplify this expression as much as possible.
  6. Take the Limit (for Infinite Series). If the series is infinite, compute limN→∞ SN. If this limit exists and is finite (L), then the series converges to L. If the limit does not exist or is infinite, the series diverges.

Detailed Examples with Different Forms

Let’s apply the steps to various series to solidify your understanding.

Example 1: Simple Linear Factors

Find the sum of ∑n=1∞ 1/(n(n+2)).

  1. Partial Fractions: We want constants A and B such that: 1/(n(n+2)) = A/n + B/(n+2). Solving gives A = 1/2, B = -1/2. So, term = (1/2)[1/n – 1/(n+2)].
  2. Write Partial Sum:
    SN = (1/2)[ (1 – 1/3) + (1/2 – 1/4) + (1/3 – 1/5) + (1/4 – 1/6) + … + (1/(N-1) – 1/(N+1)) + (1/N – 1/(N+2)) ].
  3. Observe Cancellation: The negative term -1/3 cancels with a later +1/3. Similarly, -1/4 cancels with a later +1/4. This pattern continues. What survives? The positive terms 1 and 1/2, and the negative terms -1/(N+1) and -1/(N+2).
  4. Simplify: SN = (1/2)[1 + 1/2 – 1/(N+1) – 1/(N+2)] = (1/2)[3/2 – (2N+3)/((N+1)(N+2))].
  5. Take Limit: As N→∞, 1/(N+1) and 1/(N+2) go to 0. So, Sum = (1/2)(3/2) = 3/4.

Example 2: Series with Square Roots

Telescoping isn’t limited to rational functions. Consider ∑n=1∞ (√(n+1) – √n).

  1. It’s already in telescoping form! Here, an = bn – bn+1, where bn = √n.
  2. Write Partial Sum: SN = (√2 – √1) + (√3 – √2) + (√4 – √3) + … + (√(N+1) – √N).
  3. Observe Cancellation: Every term cancels except -√1 and +√(N+1).
  4. Simplify: SN = √(N+1) – 1.
  5. Take Limit: limN→∞ SN = ∞. This series diverges to infinity because the partial sums grow without bound.

Example 3: Slightly More Complex Fractions

Find the sum of ∑n=1∞ 2/(4n² – 1).

  1. Partial Fractions: Factor denominator: (2n-1)(2n+1). Set 2/((2n-1)(2n+1)) = A/(2n-1) + B/(2n+1). Solving yields A=1, B=-1. So, term = 1/(2n-1) – 1/(2n+1).
  2. Write Partial Sum: SN = (1/1 – 1/3) + (1/3 – 1/5) + (1/5 – 1/7) + … + (1/(2N-1) – 1/(2N+1)).
  3. Observe Cancellation: All terms cancel except the first positive 1/1 and the last negative -1/(2N+1).
  4. Simplify: SN = 1 – 1/(2N+1).
  5. Take Limit: limN→∞ (1 – 1/(2N+1)) = 1. So the series sums to 1.

Common Pitfalls and How to Avoid Them

Even with a clear method, mistakes can happen. Here are common errors and how to sidestep them.

  • Not writing enough terms. The cancellation pattern might not be obvious if you only write 2-3 terms. Write at least 4 or 5, plus the generic last term, to see the full pattern.
  • Incorrect partial fractions. Double-check your algebra here. A small sign error will ruin the entire cancellation. Plug in a couple values of n to verify your decomposition is correct.
  • Forgetting the leading coefficient. In examples like 1/(n(n+2)), we factored out a 1/2. If you forget it, your final answer will be off by a factor of 2.
  • Misidentifying the surviving terms. In series where the cancellation isn’t strictly between term n and n+1 (like the n(n+2) example), carefully track which positives and negatives remain. Writing terms in a column is essential.
  • Assuming convergence. Not all telescoping series converge. Always check the limit of the remaining term bN+k. If it doesn’t approach a finite number, the series diverges.

Practice Problems for You

Try these on your own. The answers are provided below, but give it a shot first!

  1. ∑n=1∞ ln( n/(n+1) ). Hint: Use logarithm properties.
  2. ∑n=1∞ 1/(√n + √(n+1)). Hint: Rationalize the denominator.
  3. ∑n=1∞ 3/(n(n+3)).

Solutions:

  1. Term = ln(n) – ln(n+1). SN = ln(1) – ln(N+1) = -ln(N+1). Limit is -∞, so series diverges.
  2. Multiply numerator and denominator by (√(n+1) – √n) to get term = √(n+1) – √n. This telescopes to SN = √(N+1) – 1, which diverges.
  3. Partial fractions: (1/n) – 1/(n+3). SN = 1 + 1/2 + 1/3 – 1/(N+1) – 1/(N+2) – 1/(N+3). Limit is 1 + 1/2 + 1/3 = 11/6.

Why This Technique Matters in Calculus

Telescoping series are more than just a neat trick. They provide concrete examples of series convergence that you can verify by hand. This helps build intuition for more advanced tests like the Integral Test or Comparison Test.

They also appear in real applications, such as in signal processing or the analysis of algorithms, where sums over structures that simplify are common. Mastering this foundational skill makes learning subsequent topics easier.

Furthermore, the process of partial fraction decomposition is a key skill in its own right, used extensively in integral calculus. So you’re learning two important techniques at once.

Advanced Variations

Sometimes, the telescoping isn’t immediate. You might need to use a creative algebraic manipulation.

When the Term is a Product

For a series like ∑ n an, you might try expressing nan as a difference. Other times, taking the logarithm of a product, as in the practice problem, reveals the telescoping nature.

Telescoping in Both Directions

Rarely, you might encounter a series where terms cancel in a more complex “fold.” The principle remains the same: write terms, look for subtraction patterns, and simplify what’s left.

Connecting to Partial Sums and Convergence Tests

The Nth partial sum, SN, is the key object of study for any series. For a telescoping series, we find an explicit, closed-form formula for SN. This is the gold standard—most series tests only tell you if a sum exists, not what it is.

Because we have a formula, the convergence test reduces to checking if limN→∞ SN exists. This direct approach bypasses the need for tests like the Ratio Test, providing both a yes/no answer and the actual sum value.

Final Checklist for Your Work

Before you declare victory, run through this list:

  • ✓ Is the partial fraction decomposition 100% correct?
  • ✓ Did I write out at least 4-5 specific terms plus the generic last term?
  • ✓ Did I correctly identify all surviving terms after cancellation?
  • ✓ Is my formula for SN simplified?
  • ✓ For an infinite series, did I take the limit of SN correctly?
  • ✓ Does my final answer make sense? (e.g., a positive series shouldn’t sum to a negative number).

FAQ Section

What is a telescoping series?
A telescoping series is one where consecutive terms cancel each other out when you write out the partial sum. This leaves only a few terms from the beginning and end of the sum, making it easy to compute the total.

How do you know if a series is telescoping?
Look for series whose general term can be rewritten as a difference of two expressions, like bn – bn+1. Series with products in the denominator (e.g., n(n+1)) are classic candidates and often require partial fractions to see the telescoping form.

Can all series be made to telescope?
No, definitely not. Telescoping is a specific property of the series structure. Many common series, like geometric series or p-series, do not telescope. However, the technique is very effective for the family of series that do have this property.

Do telescoping series always converge?
Not always. Convergence depends on the limit of the remaining term after cancellation. If limN→∞ bN+1 is a finite number, the series converges. If this limit is infinite or does not exist, the series diverges, as seen in the square root example.

What’s the difference between a telescoping sum and a geometric sum?
They are fundamentally different. A geometric series has a constant ratio between terms. A telescoping series relies on cancellation of consecutive terms. Their methods of solution are completely distinct, though both can result in a simple formula for the sum.

Is partial fractions the only way to find a telescoping sum?
It’s the most common algebraic tool, but not the only one. Other methods include rationalizing denominators (for series with roots) or using logarithmic and trigonometric identities to create a difference form.

In conclusion, learning how to find sum of telescoping series is about recognizing a pattern and executing a clear, step-by-step plan. Start by decomposing the term, write out the partial sum to see the cancellation, simplify, and take the limit. With practice, you’ll be able to identify these series quickly and compute their sums accurately, turning a seemingly long sum into a simple calculation.