More on Sequences — Question 4

*Question 4 Let $a_n=\sum_{k=1}^n2^{-k}$.

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

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

3. Find the exact remaining gap $L-a_n$ and the smallest $n$ for which this gap is below $10^{-3}$.

*Question 4 – Solution

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

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

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

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