Find the sum of \(n\) terms of the geometric progression \[\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\dotsb.\]

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 \[S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\dotsb+\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}.\]

Multiplying the above by \(\dfrac{1}{3}\), we have \[\begin{align*} \frac{1}{3}S_n&=\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\dotsb+\frac{1}{2}\left(\frac{1}{3}\right)^n,\\ \end{align*}\] and now subtracting this equation from the one above it, since all the fractions \(\dfrac{1}{6}, \dfrac{1}{18}, \dotsc, \dfrac{1}{2}\left(\dfrac{1}{3}\right)^{n-1}\) cancel, \[\begin{align*} S_n-\frac{1}{3}S_n &= \frac{1}{2}-\frac{1}{2}\left(\frac{1}{3}\right)^n\\ \Longrightarrow \quad \frac{2}{3}S_n &= \frac{1}{2}\left[1-\left(\frac{1}{3}\right)^n\right] \left(\text{ taking out a factor of } \frac{1}{2}\right),\\ \Longrightarrow \quad S_n &= \frac{3}{4}\left[1-\left(\frac{1}{3}\right)^n\right] \left(\text{ dividing both sides by }\frac{2}{3}\right). \end{align*}\]

Deduce the sum to infinity of this series.

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

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

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 \(S_8\).