When does the sum of this series first exceed $2999/4000$?

When does the sum of this series first exceed $2999/4000$?

Question

Find the sum of $n$ terms of the geometric progression $\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\dots b.$ [If a general formula is used, it must first be proved.]

Deduce the sum to infinity of this series.

Find the least number of terms of the series which must be taken for their sum to exceed $\dfrac{2999}{4000}$.

Suggestion

What is the common ratio of this geometric series?

What is the $n$th term of this geometric series?

Let $S_n$ denote the sum of the first $n$ terms. Could we write the expansion of $\frac{1}{3}S_n$ underneath the expression for $S_n$?

What do we notice? What happens if we take the two expressions away from each other?

When $n\to \infty$, what does $S_n$ tend towards?

Could we try to find an $n$ such that $S_n = \dfrac{2999}{4000}$?

Could we write down the first few powers of $3$?

Solution

This is the sum of a geometric series with first term $a=\dfrac{1}{2}$ and common ratio $r=\dfrac{1}{3}$. The $n$th term in this series is $ar^{n-1}=\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}.$

Let the sum of the first $n$ terms be $S_n$, so $S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\dots b+\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}.$

Multiplying the above by

$\dfrac{1}{3}$

, we have

$$\begin{align*} \frac{1}{3}S_n&=\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\dots b+\frac{1}{2}\left(\frac{1}{3}\right)^n,\\ \end{align*}$$

and now subtracting this equation from the one above it, since all the fractions

$\dfrac{1}{6}, \dfrac{1}{18}, \dots c, \dfrac{1}{2}\left(\dfrac{1}{3}\right)^{n-1}$

cancel,

$$\begin{align*} S_n-\frac{1}{3}S_n &= \frac{1}{2}-\frac{1}{2}\left(\frac{1}{3}\right)^n\\ \Longrightarrow \quad \frac{2}{3}S_n &= \frac{1}{2}\left[1-\left(\frac{1}{3}\right)^n\right] \left(\text{ taking out a factor of } \frac{1}{2}\right),\\ \Longrightarrow \quad S_n &= \frac{3}{4}\left[1-\left(\frac{1}{3}\right)^n\right] \left(\text{ dividing both sides by }\frac{2}{3}\right). \end{align*}$$

As $n\to\infty$, $\left(\frac{1}{3}\right)^n \to 0.$ Therefore, the sum to infinity of this geometric series is $S_{\infty}=\frac{3}{4}\left(1-0\right) = \frac{3}{4}.$

We are looking for the $n$ such that $S_{n-1}\leq\frac{2999}{4000}\;\; \text{and}\;\; S_n>\frac{2999}{4000}.$

So we want to solve for

$n$

$$\begin{align*} \frac{3}{4}\left[1-\left(\frac{1}{3}\right)^{n}\right]&=\frac{2999}{4000}=\frac{3000-1}{4000} \end{align*}$$

dividing through by

$\dfrac{3}{4}$

gives

$$\begin{align*} 1-\left(\frac{1}{3}\right)^{n}&=\frac{3000-1}{3000} \end{align*}$$

and this is the same as

$$\begin{align*} 1-\frac{1}{3^{n}}&= 1-\frac{1}{3000}\\ \Longrightarrow \quad \frac{1}{3000}&= \frac{1}{3^{n}} \\ \Longrightarrow \quad 3^{n} &= 3000 \\ \Longrightarrow \quad 3^{n-1} &= 1000. \end{align*}$$

Taking logs here gives $n = 1 + \dfrac{\log(1000)}{\log 3} \approx 7.29.$ So the first $S_n$ to exceed $\dfrac{2999}{4000}$ is $S_8$.