Calculus II · Unit 4A · lesson

Telescoping Series

Expose cancellation through finite partial sums, track surviving boundary terms, and evaluate one-step and shifted telescoping series.

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Section overview

Infinite series and foundational examples

What this section is building

Expose cancellation through finite partial sums, track surviving boundary terms, and evaluate one-step and shifted telescoping series.

Notice

A series converges exactly when its sequence of partial sums approaches a finite value.

Decide

Check the term limit first, then look for an exact partial-sum pattern before selecting a comparison test.

Avoid

Concluding convergence from terms approaching zero or canceling inside an unwritten infinite expression.

Concept

Learning objectives

use partial fractions or algebraic decomposition to expose cancellation and compute a series sum.

Telescoping Series

Explanation

Write the partial sum before trusting the cancellation

A telescoping series contains terms arranged so that most contributions cancel in a finite partial sum. The cancellation is not visible from the infinite notation alone; it appears after several terms are written with their signs. This is why a correct solution displays $s_N$ rather than announcing that "everything cancels."

After cancellation, a few boundary terms remain. The series converges if those remaining terms approach a finite limit. Partial fractions frequently create telescoping structure, especially for rational terms involving consecutive factors such as $n(n+1)$ or $(n+a)(n+b)$ .

Bridge

Cancellation is a fact about partial sums

A telescoping series is not a mysterious new species of infinite sum. It is an ordinary series whose finite partial sums simplify because most terms cancel. The safe procedure is therefore to write several terms of $s_N$ , mark the cancellation, and keep the boundary terms that survive. Only after the finite formula is correct do we let $N\to\infty$ .

Partial fractions often reveal the hidden cancellation. The important question is not merely whether adjacent-looking symbols resemble one another, but whether the shifted indices line up across the entire finite sum. A gap of two or three indices may leave several terms at each boundary rather than just one.

**Interior terms cancel; boundary terms survive**
A finite telescoping partial sum is displayed as shifted positive and negative rows. Matching interior terms cancel, leaving only terms at the two ends.
Interior terms cancel; boundary terms survive. A cancellation chain showing interior terms disappearing while boundary terms remain.

Proof idea

The limit enters only after the algebra is finite

For every fixed $N$ , cancellation is ordinary finite algebra. Once $s_N$ has been reduced to its surviving terms, convergence is determined by the limit of that finite formula.

Concept

Telescoping workflow

Decompose the term, write a finite partial sum, cancel only terms actually present, simplify the boundary expression, and then let $N\to\infty$ .

Guided walkthrough

A standard telescoping sum

Since

\frac1{n(n+1)}=\frac1n-\frac1{n+1},

we have

s_N=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1N-\frac1{N+1}\right)=1-\frac1{N+1}.

Thus the infinite series sums to $1$ .

Worked example

Telescoping with a two-step shift

Evaluate

\sum_{n=1}^{\infty}\frac{1}{n(n+2)}.

First decompose

\frac{1}{n(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right).

Then

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

The final two fractions vanish as $N\to\infty$ , so

\sum_{n=1}^{\infty}\frac{1}{n(n+2)}=\frac12\left(1+\frac12\right)=\frac34.

Notice that two initial terms survive because the shift is two.

Common mistake

Never cancel inside an unwritten infinite expression

Cancellation must be demonstrated in the finite partial sum. Informally crossing out infinitely many terms can hide boundary terms or invalid rearrangements.

Interactive checku4a-telescoping_series-01

Evaluate $\sum_{n=1}^{\infty}\frac1{n(n+1)}$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Use $1/[n(n+1)]=1/n-1/(n+1)$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

Evaluate $\sum_{n=1}^{\infty}2/[(n+1)(n+3)]$ .

Exercise

Write the first five terms before canceling.

Exercise

Give a telescoping series that diverges because a boundary term fails to settle.

Exercise

Explain why cancellation in an infinite expression must be justified through finite partial sums.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Telescoping Series

Explanation

Write the partial sum before trusting the cancellation

Bridge

Cancellation is a fact about partial sums

Read this graph as text

Interior terms cancel; boundary terms survive. A finite telescoping partial sum is displayed as shifted positive and negative rows. Matching interior terms cancel, leaving only terms at the two ends. A cancellation chain showing interior terms disappearing while boundary terms remain. Preserve the misconception control described in the lesson and storyboard.

Written labels, distinct line styles, markers, and fill patterns communicate every relationship in interior terms cancel; boundary terms survive; color is never the only cue.

Why it matters: A cancellation chain showing interior terms disappearing while boundary terms remain.

Interior terms cancel; boundary terms survive

A finite telescoping partial sum is displayed as shifted positive and negative rows. Matching interior terms cancel, leaving only terms at the two ends.

Interior terms cancel; boundary terms survive. A cancellation chain showing interior terms disappearing while boundary terms remain.

Proof idea

The limit enters only after the algebra is finite

For every fixed $N$ , cancellation is ordinary finite algebra. Once $s_N$ has been reduced to its surviving terms, convergence is determined by the limit of that finite formula.

Concept

Telescoping workflow

Decompose the term, write a finite partial sum, cancel only terms actually present, simplify the boundary expression, and then let $N\to\infty$ .

Guided walkthrough

A standard telescoping sum

Since

\frac1{n(n+1)}=\frac1n-\frac1{n+1},

we have

s_N=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1N-\frac1{N+1}\right)=1-\frac1{N+1}.

Thus the infinite series sums to $1$ .

Worked example

Telescoping with a two-step shift

Evaluate

\sum_{n=1}^{\infty}\frac{1}{n(n+2)}.

First decompose

\frac{1}{n(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right).

Then

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

The final two fractions vanish as $N\to\infty$ , so

\sum_{n=1}^{\infty}\frac{1}{n(n+2)}=\frac12\left(1+\frac12\right)=\frac34.

Notice that two initial terms survive because the shift is two.

Common mistake

Never cancel inside an unwritten infinite expression

Cancellation must be demonstrated in the finite partial sum. Informally crossing out infinitely many terms can hide boundary terms or invalid rearrangements.

Interactive checku4a-telescoping_series-01

Evaluate $\sum_{n=1}^{\infty}\frac1{n(n+1)}$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Use $1/[n(n+1)]=1/n-1/(n+1)$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

Evaluate $\sum_{n=1}^{\infty}2/[(n+1)(n+3)]$ .

Exercise

Write the first five terms before canceling.

Exercise

Give a telescoping series that diverges because a boundary term fails to settle.

Exercise

Explain why cancellation in an infinite expression must be justified through finite partial sums.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Telescoping Series

Related resources

Original instruction with traceable references.

Telescoping Series

Related resources

Original instruction with traceable references.