Sequences — Question 3

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Question 3

Let an=nn+1a_n=\dfrac{n}{n+1} for n≥1n\ge1.

  1. Prove algebraically (without derivatives) that (an)(a_n) is strictly increasing.

  2. Find an upper bound and use the Monotone Convergence Theorem to explain why a limit exists.

  3. Find the limit and express the error |an−L||a_n-L| exactly.

Original worksheet page 1: question and worked solution for 4-1-003
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Question 3 – Solution

Step 1: Prove monotonicity algebraically.

Compare consecutive terms by subtraction: an+1−an=n+1n+2−nn+1=1(n+1)(n+2)>0.a_{n+1}-a_n=\frac{n+1}{n+2}-\frac n{n+1}=\frac1{(n+1)(n+2)}>0. The denominator is positive for n≥1n\ge1, so an+1−an>0a_{n+1}-a_n>0 and (an)(a_n) is strictly increasing.

Step 2: Find an upper bound.

Rewrite the term as an=1−1n+1<1,a_n=1-\frac1{n+1}<1, Thus an<1a_n<1 for every nn, so the sequence is bounded above by 11.

Step 3: Establish convergence and calculate the limit.

An increasing sequence bounded above converges by the Monotone Convergence Theorem. Using the rewritten formula, limn→∞an=limn→∞(1−1n+1)=1.\lim_{n\to\infty}a_n=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)=1.

Step 4: Express the error exactly.

Since an<1a_n<1, the exact error is |an−1|=1−an=1/(n+1)|a_n-1|=1-a_n=1/(n+1), which tends to zero.

Original worksheet page 2: question and worked solution for 4-1-003

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