Series - The Basics — Question 1

Question 1 – Solution

Step 1: Identify the geometric structure.

Writing the first terms as $3/4+3/16+3/64+\cdots$ shows that the first term is $a=3/4$ and the common ratio is $r=1/4$. Since $|r|<1$, the infinite geometric-series formula applies: $$S=\frac{a}{1-r}=\frac{3/4}{1-1/4}=1.$$

Step 2: Confirm the result from partial sums.

For a geometric series, the finite sum is $s_N=a(1-r^N)/(1-r)$. Substitution gives $$s_N=\frac{3}{4}\frac{1-(1/4)^N}{1-1/4}=1-4^{-N},$$ Since $4^{-N}\to0$, $s_N\to1$, confirming the result from the definition of an infinite series.

Step 3: Find and use the exact remainder.

The exact remainder after $N$ terms is $$R_N=S-s_N=4^{-N}.$$ We require $4^{-N}<10^{-4}$, equivalently $4^N>10^4$. Since $4^6=4096\le10^4$ but $4^7=16384>10^4$, the smallest choice is $N=7$. The strict inequality is important.