More on Sequences — Question 4

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Question 4

Let an=∑k=1n2−ka_n=\sum_{k=1}^n2^{-k}.

  1. Derive a closed formula for ana_n from the finite geometric-sum formula.

  2. Prove that (an)(a_n) converges and find its limit.

  3. Find the exact remaining gap L−anL-a_n and the smallest nn for which this gap is below 10−310^{-3}.

Original worksheet page 1: question and worked solution for 4-2-004
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Question 4 – Solution

Step 1: Identify and sum the finite geometric series.

The terms form a geometric sequence with first term a=1/2a=1/2 and common ratio r=1/2r=1/2. Therefore an=(1/2)(1−(1/2)n)1−1/2=1−2−n.a_n=\frac{(1/2)(1-(1/2)^n)}{1-1/2}=1-2^{-n}.

Step 2: Determine convergence and the limit.

Since 2−n→02^{-n}\to0, the closed formula gives an→L=1a_n\to L=1. Also an+1−an=2−(n+1)>0a_{n+1}-a_n=2^{-(n+1)}>0 and an<1a_n<1, so the sequence is increasing and bounded above, independently confirming convergence by the Monotone Convergence Theorem.

Step 3: Compute the exact remaining gap.

The exact error after nn terms is L−an=2−n.L-a_n=2^{-n}.

Step 4: Solve the accuracy inequality.

We need 2−n<10−32^{-n}<10^{-3}, equivalently 2n>10002^n>1000. Since 29=512≤10002^9=512\le1000 but 210=1024>10002^{10}=1024>1000, the smallest qualifying index is n=10n=10.

Original worksheet page 2: question and worked solution for 4-2-004

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