Series - The Basics — Question 1

*Question 1 Consider the series $\displaystyle\sum_{n=1}^{\infty}\frac3{4^n}$.

1. Identify the first term and common ratio, then evaluate the series using the infinite geometric-series formula.

2. Derive a formula for the $N$th partial sum $s_N$ and evaluate $\lim_{N\to\infty}s_N$.

3. Find the exact remainder $R_N=S-s_N$ and determine the smallest $N$ for which $R_N<10^{-4}$.

*Question 1 – Solution

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: $$S=\frac{a}{1-r}=\frac{3/4}{1-1/4}=1.$$

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.

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.