Infinite series: the range of x, r and the first term Problem set A Worked solutions

Find the sum to infinity of each series.

1. (a) \(\frac23-\frac29+\frac{2}{27}-\dots\)

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

2. (b) \(24-18+\frac{27}{2}-\dots\)

   • \(u_1=24,\ r=-\frac{3}{4}\)
   • \(S_\infty=\frac{24}{1-(-\frac{3}{4})}\)
   • \(S_\infty=\frac{96}{7}\)

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

1. (a) \((x-6)+(x-6)^2+(x-6)^3+\dots\)

   • Find r from the first two terms, then \(|r|<1\): \(-1<x-6<1\)
   • Add 6 to every part.
   • \(5<x<7\)

2. (b) \(1+\frac x4+\frac{x^2}{16}+\dots\)

   • Find r from the first two terms, then \(|r|<1\): \(-1<\frac x4<1\)
   • Multiply every part by 4.
   • \(-4<x<4\)

3. (c) \(1-\frac{5x}{3}+\frac{25x^2}{9}-\dots\)

   • Find r from the first two terms, then \(|r|<1\): \(-1<-\frac{5x}{3}<1\)
   • Multiply by \(-\frac35\) and reverse the signs.
   • \(-0.6<x<0.6\)

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

1. (a) \(1+5x^2+25x^4+\dots\)

   • \(r=5x^2\), which is never negative
   • \(5x^2<1\Rightarrow x^2<\frac15\)
   • \(-0.447<x<0.447\)

2. (b) \(1+\frac2x+\frac4{x^2}+\dots\)

   • \(r=\frac2{x}\)
   • \(\left|\frac2x\right|<1\Rightarrow |x|>2\) (multiply by \(|x|\))
   • \(x<-2 \text{ or } x>2\)

A geometric series has first term 15 and sum to infinity 40. Find the common ratio.

• \(40=\frac{15}{1-r}\Rightarrow 1-r=\frac{15}{40}=0.375\)
• \(r=0.625\)
• \(|r|<1\), so this is allowed

A geometric series has sum to infinity 35 and common ratio \(-\frac{2}{5}\). Find the first term, and write down the first three terms.

• \(u_1=S_\infty(1-r)=35\left(1-\left(-\frac{2}{5}\right)\right)\)
• \(u_1=49\)
• Multiply by \(-\frac{2}{5}\): 49, −19.6, 7.84

A series is \(\sum_{k=1}^{\infty}12\left(\frac13\right)^k\).

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

   • Put in \(k=1,2,3\)
   • First term: \(12\times\frac{1}{3}=4\)
   • Terms: \(4\), \(\frac{4}{3}\), \(\frac{4}{9}\)

2. (b) Find the sum to infinity.

   • \(u_1=4,\ r=\frac{1}{3}\)
   • \(S_\infty=\frac{4}{1-\frac{1}{3}}\)
   • \(S_\infty=6\)

A geometric series has second term 8 and sum to infinity 50.

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

   • \(u_1=\frac{8}{r}\), so \(8=50r(1-r)\)
   • \(50r^2-50r+8=0\)
   • From the calculator: \(r=\frac{1}{5}\) or \(r=\frac{4}{5}\)

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

   • \(u_1=\frac{8}{r}\)
   • \(r=\frac{1}{5}\) gives \(u_1=40\)
   • \(r=\frac{4}{5}\) gives \(u_1=10\)

A series is \(4+4\left(\frac{x-1}{3}\right)+4\left(\frac{x-1}{3}\right)^2+\dots\)

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

   • \(r=\frac{x-1}{3}\)
   • \(-1<\frac{x-1}{3}<1\)
   • Multiply by 3: \(-3<x-1<3\). Add 1.
   • \(-2<x<4\)

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

   • \(S_\infty=\frac{4}{1-\frac{x-1}{3}}\)
   • \(=\frac{12}{4-x}\)

3. (c) Find \(S_\infty\) when \(x=2\).

   • \(x=2\) is inside the range
   • \(S_\infty=\frac{12}{4-2}=6\)

The sum to infinity of \(\sum_{k=1}^{\infty}5\left(\frac x4\right)^{k-1}\) is 8. Find the value of x.

• \(u_1=5,\ r=\frac x4\)
• \(\frac{5}{1-\frac x4}=8\Rightarrow 1-\frac x4=\frac58\)
• \(\frac x4=\frac38\Rightarrow x=1.5\)
• \(\left|\frac{1.5}{4}\right|<1\), so x is in the range

A trust will pay out a total of 50 000 euros over an unlimited number of years. It pays 12 000 euros in the first year, and each later year pays a constant fraction r of the year before. Find r.

• \(S_\infty=\frac{u_1}{1-r}\Rightarrow 1-r=\frac{12}{50}\)
• \(r=0.76\)

The weekly water supplied by a well forms a geometric sequence. Over all the weeks to come it will supply 54 thousand litres, and in the first three weeks it supplies 38 thousand litres. Find the volume, in thousand litres, supplied in the first week.

• \(S_3=S_\infty(1-r^3)\Rightarrow 38=54(1-r^3)\)
• \(r^3=1-\frac{38}{54}=\frac{8}{27}\), so \(r=\frac{2}{3}\)
• \(u_1=54(1-\frac{2}{3})=18\)

A geometric series has second term \(-5\) and sum to infinity 16.

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

   • \(-5=16r(1-r)\Rightarrow 16r^2-16r-5=0\)
   • From the calculator: \(r=-\frac{1}{4}\) or \(r=\frac{5}{4}\)
   • \(\left|\frac{5}{4}\right|>1\), so reject it

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

   • \(u_1=\frac{-5}{-\frac{1}{4}}=20\)
   • Check: \(\frac{20}{1-(-\frac{1}{4})}=16\)