Infinite series: the range of x, r and the first term Problem set B One question at a time

Find the sum to infinity of each series.

1. (a) \(\frac45-\frac{4}{15}+\frac{4}{45}-\dots\)

   • \(u_1=\frac{4}{5},\ r=-\frac{1}{3}\)
   • \(S_\infty=\frac{\frac{4}{5}}{1-(-\frac{1}{3})}\)
   • \(S_\infty=\frac{3}{5}\)

2. (b) \(30-20+\frac{40}{3}-\dots\)

   • \(u_1=30,\ r=-\frac{2}{3}\)
   • \(S_\infty=\frac{30}{1-(-\frac{2}{3})}\)
   • \(S_\infty=18\)

Find the range of values of x for which each series converges.

1. (a) \((x+7)+(x+7)^2+(x+7)^3+\dots\)

   • Find r from the first two terms, then \(|r|<1\): \(-1<x+7<1\)
   • Subtract 7 from every part.
   • \(-8<x<-6\)

2. (b) \(1+\frac x6+\frac{x^2}{36}+\dots\)

   • Find r from the first two terms, then \(|r|<1\): \(-1<\frac x6<1\)
   • Multiply every part by 6.
   • \(-6<x<6\)

3. (c) \(1-\frac{4x}{5}+\frac{16x^2}{25}-\dots\)

   • Find r from the first two terms, then \(|r|<1\): \(-1<-\frac{4x}{5}<1\)
   • Multiply by \(-\frac54\) and reverse the signs.
   • \(-1.25<x<1.25\)

Find the range of values of x for which each series converges.

1. (a) \(1+6x^2+36x^4+\dots\)

   • \(r=6x^2\), which is never negative
   • \(6x^2<1\Rightarrow x^2<\frac16\)
   • \(-0.408<x<0.408\)

2. (b) \(1+\frac3x+\frac9{x^2}+\dots\)

   • \(r=\frac3{x}\)
   • \(\left|\frac3x\right|<1\Rightarrow |x|>3\) (multiply by \(|x|\))
   • \(x<-3 \text{ or } x>3\)

A geometric series has first term 14 and sum to infinity 35. Find the common ratio.

• \(35=\frac{14}{1-r}\Rightarrow 1-r=\frac{14}{35}=0.4\)
• \(r=0.6\)
• \(|r|<1\), so this is allowed

A geometric series has sum to infinity 60 and common ratio \(-\frac{1}{4}\). Find the first term, and write down the first three terms.

• \(u_1=S_\infty(1-r)=60\left(1-\left(-\frac{1}{4}\right)\right)\)
• \(u_1=75\)
• Multiply by \(-\frac{1}{4}\): 75, −18.8, 4.69

A series is \(\sum_{k=1}^{\infty}15\left(\frac25\right)^k\).

1. (a) Write down the first three terms.

   • Put in \(k=1,2,3\)
   • First term: \(15\times\frac{2}{5}=6\)
   • Terms: \(6\), \(\frac{12}{5}\), \(\frac{24}{25}\)

2. (b) Find the sum to infinity.

   • \(u_1=6,\ r=\frac{2}{5}\)
   • \(S_\infty=\frac{6}{1-\frac{2}{5}}\)
   • \(S_\infty=10\)

A geometric series has second term 6 and sum to infinity 27.

1. (a) Find the two possible values of r.

   • \(u_1=\frac{6}{r}\), so \(6=27r(1-r)\)
   • \(27r^2-27r+6=0\)
   • From the calculator: \(r=\frac{1}{3}\) or \(r=\frac{2}{3}\)

2. (b) Find the two possible values of \(u_1\).

   • \(u_1=\frac{6}{r}\)
   • \(r=\frac{1}{3}\) gives \(u_1=18\)
   • \(r=\frac{2}{3}\) gives \(u_1=9\)

A series is \(6+6\left(\frac{x+2}{5}\right)+6\left(\frac{x+2}{5}\right)^2+\dots\)

1. (a) Find the range of values of x for which the series converges.

   • \(r=\frac{x+2}{5}\)
   • \(-1<\frac{x+2}{5}<1\)
   • Multiply by 5: \(-5<x+2<5\). Subtract 2.
   • \(-7<x<3\)

2. (b) Find an expression for \(S_\infty\) in terms of x.

   • \(S_\infty=\frac{6}{1-\frac{x+2}{5}}\)
   • \(=\frac{30}{3-x}\)

3. (c) Find \(S_\infty\) when \(x=1\).

   • \(x=1\) is inside the range
   • \(S_\infty=\frac{30}{3-1}=15\)

The sum to infinity of \(\sum_{k=0}^{\infty}6\left(\frac x5\right)^{k}\) is 10. Find the value of x.

• \(u_1=6,\ r=\frac x5\)
• \(\frac{6}{1-\frac x5}=10\Rightarrow 1-\frac x5=\frac35\)
• \(\frac x5=\frac25\Rightarrow x=2\)
• \(\left|\frac25\right|<1\), so x is in the range

A crew is cleaning an oil spill. In total it will remove 900 cubic metres of oil if it works for ever. It removes 270 cubic metres on the first day, and each later day removes a constant fraction r of the day before. Find r.

• \(S_\infty=\frac{u_1}{1-r}\Rightarrow 1-r=\frac{270}{900}\)
• \(r=0.7\)

The yearly ore output of a mine forms a geometric sequence. Over all the years to come it will produce 81 thousand tonnes, and in the first three years it produces 78 thousand tonnes. Find the output, in thousand tonnes, in the first year.

• \(S_3=S_\infty(1-r^3)\Rightarrow 78=81(1-r^3)\)
• \(r^3=1-\frac{78}{81}=\frac{1}{27}\), so \(r=\frac{1}{3}\)
• \(u_1=81(1-\frac{1}{3})=54\)

A geometric series has second term \(-6\) and sum to infinity 25.

1. (a) Find the possible values of r. Reject any that cannot be used.

   • \(-6=25r(1-r)\Rightarrow 25r^2-25r-6=0\)
   • From the calculator: \(r=-\frac{1}{5}\) or \(r=\frac{6}{5}\)
   • \(\left|\frac{6}{5}\right|>1\), so reject it

2. (b) Find \(u_1\).

   • \(u_1=\frac{-6}{-\frac{1}{5}}=30\)
   • Check: \(\frac{30}{1-(-\frac{1}{5})}=25\)