Sequences and Series

• \(3, 5, 7, 9, 11 \dots\) is a sequence.
• Each number in \(3, 5, 7, 9, 11, \dots\) is called a term of the sequence.
• We see
  • the 1st term is \(a_1 = 3 = 1 \cdot 2 + 1\)
  • the 2nd term is \(a_2 = 5 = 2 \cdot 2 + 1\)
  • the 3rd term is \(a_3 = 7 = 3 \cdot 2 + 1\)
  • the 4rd term is \(a_4 = 9 = 4 \cdot 2 + 1\)
  • the 5rd term is \(a_5 = 11 = 5 \cdot 2 + 1\)
  • \(\dots\)
  • the \(i\)th term is \(a_i = f(i) = i \cdot 2 + 1\)
  • so the general term of the sequence is \(a_i = f(i) = i \cdot 2 + 1\).

Suppose \(a_1, a_2, a_3, a_4, a_5, \dots\) is a sequence. If \(\frac{a_{i+1}}{a_i}\) equals the same constant \(r\) for all \(i = 1, 2, 3, \dots\), then the sequence is called a geometric sequence.

• Example 1

  Suppose a sequence \(a_1 = 3, a_2 = 9, a_3 = 27, a_4 = 81, a_5 = 243, \dots\) is given. We see that

  • \(\frac{a_2}{a_1} = \frac{9}{3} = 3\)
  • \(\frac{a_3}{a_2} = \frac{27}{9} = 3\)
  • \(\frac{a_4}{a_3} = \frac{81}{27} = 3\)
  • \(\frac{a_5}{a_4} = \frac{243}{81} = 3\)
  • \(\dots\)

  That is, \(\frac{a_{i+1}}{a_i} = 3\) for all \(i = 1, 2, 3, \dots\), so this sequence is a geometric sequence.

• Example 2

  Suppose a sequence \(a_1 = 3, a_2 = 9, a_3 = 21, \dots\) is given. We see that

  • \(\frac{a_2}{a_1} = \frac{9}{3} = 3\)
  • \(\frac{a_3}{a_2} = \frac{21}{9} = \frac{7}{3} \neq 3\)

  Since \(\frac{a_2}{a_1} \neq \frac{a_3}{a_2}\), this sequence is NOT a geometric sequence.

For a geometric sequence \(a_1, a_2, a_3, a_4, a_5, \dots\), the constant \(r = \frac{a_{i+1}}{a_i}\) is called the common ratio.

Remark:

• \(r\) is called the common ratio because it is the same between any two consecutive terms.
• For a geometric sequence, \(r = \frac{a_{i+1}}{a_i}\) is the same for all \(i = 1, 2, 3, \dots\), so the value of \(r\) can be cauculated simply by \(r = \frac{a_2}{a_1}\).

If a geometric sequence has first term \(a_1\) and common ratio \(r\), then the general term of the sequence is \[ \boxed{a_i = a_1 r^{i-1}} \text{ for } i = 1,2,3,\dots .\]

Remark: Clearly

• \(a_2 = a_1 \cdot r = a_1 r^{2-1}\)
• \(a_3 = a_2 \cdot r = (a_1 \cdot r) \cdot r = a_1 r^2 = a_1 r^{3-1}\)
• \(a_4 = a_3 \cdot r = [a_2 \cdot r] \cdot r = [(a_1 \cdot r) \cdot r] \cdot r = a_1 r^3 = a_1 r^{4-1}\)
• \(\dots\)

If a geometric sequence has first term \(a_1\) and common ratio \(r\), then the sum of the first \(n\) terms, denoted by \(S_n\), is given by \[ \boxed{ S_n = \sum_{i=1}^{n} a_i = a_1 + a_2 + a_3 + \dots + a_n = \frac{a_1(r^n - 1)}{r - 1} \text{ where } r\neq 1 } \]

If a geometric sequence has first term \(a_1\) and common ratio \(r\), then the corresponding geometric series \(\sum_{i=1}^{\infty} a_i\)

• converges to \(\frac{a_1}{1-r}\) if \(r\in (-1, 1)\)
• diverges if \(r\notin (-1,1)\)

Suppose a geometric sequence \(2, -6, 18, -54, 162, \dots\) is given.

• The first term is \(a_1 = 2\)
• The common ratio is \(r = \frac{a_2}{a_1} = \frac{-6}{2} = -3\)
• The general term is \(a_n = a_1 r^{n-1} = 2 \cdot (-3)^{n-1}\)
  • The term \(a_6 = a_1 r^{6-1} = 2 \cdot (-3)^5 = -486\)
  • The term \(a_7 = a_1 r^{7-1} = 2 \cdot (-3)^6 = 1458\)
• The sum of first \(n\) terms is \(S_n = \frac{a_1(r^n - 1)}{r - 1}\) \(= \frac{2 \cdot ( (-3)^n - 1 )}{-3-1}\) \(= \frac{2 \cdot ( (-3)^n - 1 )}{-4}\) \(= \frac{(-3)^n - 1}{-2}\)
  • The sum of first \(2\) terms is \(S_2 = \frac{(-3)^2 - 1}{-2} = \frac{8}{-2} = -4\)
  • The sum of first \(5\) terms is \(S_5 = \frac{(-3)^5 - 1}{-2} = \frac{-244}{-2} = 122\)
• Since the common ratio is \(-3 \notin (-1,1)\), the series \(\sum_{i=1}^{\infty} a_n\) diverges

Suppose a geometric sequence \(2, \frac{2}{3}, \frac{2}{9}, \frac{2}{27}, \frac{2}{81} \dots\) is given.

• The first term is \(a_1 = 2\)
• The common ratio is \(r\) \(= \frac{a_2}{a_1}\) \(= \frac{\frac{2}{3}}{2}\) \(= \frac{1}{3}\)
• The general term is \(a_n = a_1 r^{n-1} = 2 \cdot \left( \frac{1}{3} \right)^{n-1}\)
  • The term \(a_6\) \(= a_1 r^{6-1}\) \(= 2 \cdot \left( \frac{1}{3} \right)^5\) \(= \frac{2}{243}\)
  • The term \(a_7\) \(= a_1 r^{7-1}\) \(= 2 \cdot \left( \frac{1}{3} \right)^6\) \(= \frac{2}{729}\)
• The sum of first \(n\) terms is \(S_n\) \(= \frac{a_1(r^n - 1)}{r - 1}\) \(= \frac{2 \cdot \left[ \left( \frac{1}{3} \right)^n - 1 \right]}{\frac{1}{3} - 1}\) \(= \frac{2 \cdot \left[ \left( \frac{1}{3} \right)^n - 1 \right]}{\frac{-2}{3}}\) \(= -3 \left[ \left( \frac{1}{3} \right)^n - 1 \right]\) \(= -3 \cdot \left( \frac{1}{3} \right)^n + 3\) \(= - \left( \frac{1}{3} \right)^{n-1} + 3\)
  • The sum of first \(2\) terms is \(S_2 = - \left( \frac{1}{3} \right)^{2-1} + 3 = \frac{8}{3}\)
  • The sum of first \(5\) terms is \(S_5 = - \left( \frac{1}{3} \right)^{5-1} + 3 = \frac{242}{81}\)
• Since the common ratio is \(\frac{1}{3} \in (-1,1)\), the series \(\sum_{i=1}^{\infty} a_n\) converges to \(\frac{a_1}{1-r} = \frac{2}{1-\frac{1}{3}} = 3\)

Find \(\sum_{i=1}^{10} \frac{5^{3i}}{2^{i-1}}\).

• Identify that this expression is the sum of first 10 terms of a sequence whose general term is \(a_i = \frac{5^{3i}}{2^{i-1}}\).

• Show that the sequence is a geometric sequence. We have \(\frac{a_{i+1}}{a_i}\) \(= \frac{\frac{5^{3(i+1)}}{2^{(i+1)-1}}}{\frac{5^{3i}}{2^{i-1}}}\) \(= \frac{5^{3(i+1)}}{2^{(i+1)-1}} \cdot \frac{2^{i-1}}{5^{3i}}\) \(= \frac{2^{i-1}}{2^{(i+1)-1}} \cdot \frac{5^{3(i+1)}}{5^{3i}}\) \(= \frac{2^{i-1}}{2^i} \cdot \frac{5^{3i+3}}{5^{3i}}\) \(= \frac{1}{2} \cdot 5^3\) \(= \frac{125}{2}\).

  That is to say, \(\frac{a_{i+1}}{a_i} = \frac{125}{2}\) is a constant for all \(i=1,2,3,\dots\). By definition, the sequence is a geometric sequence.

• The first term of the sequence is \(a_1 = \frac{5^{3\cdot 1}}{2^{1-1}} = 125\).
• The common ratio of the sequence is \(r=\frac{125}{2}\).

• Therefore, by formula, the sum of the first \(10\) terms of sequence \(a_i = \frac{5^{3i}}{2^{i-1}}\) is \(\sum_{i=1}^{10} \frac{5^{3i}}{2^{i-1}}\) \(= \frac{\left[\left(\frac{125}{2}\right)^{10} - 1\right]}{\frac{125}{2} - 1}\) \(= \frac{2\left[\left(\frac{125}{2}\right)^{10} - 1\right]}{123}\) \(= \frac{2\left(\frac{125}{2}\right)^{10} - 2}{123}\) \(= \frac{{125^{10}}{2^9} - 2}{123}\).

  You are not required to simplify it any more or write it as a decimal.

Find the first \(6\) terms of the geometric sequence whose 2nd term is \(4\) and 5th term is \(32\).

Basically, we should first find the general term of the sequence, and then calculate each one of the first \(6\) terms using the general term.

By the general term of geometric sequences \(a_i = a_1 r^{i-1}\),

• the 2nd term is \(a_2 = 4 = a_1 r^{2-1} = a_1 r \ \ (1)\)
• the 5th term is \(a_5 = 32 = a_1 r^{5-1} = a_1 r^4 \ \ (2)\)

Now we have to solve for \(a_1\) and \(r\) from the system consisting of equations \((1)\) and \((2)\). We use the substitution method.

• From \((1)\), we get \(a_1 = \frac{4}{r} \ \ (3)\).
• Put \((3)\) into \((2)\), and we get \(32 = \frac{4}{r} \cdot r^4 = 4r^3 \implies 8 = r^3 \implies r = 2 \ \ (4)\).
• Put \((4)\) into \((3)\), and we get \(a_1 = \frac{4}{2} = 2\).

Therefore, the general term of the sequence is \(a_i = a_1 r^{i-1} = 2 \cdot 2^{i-1} = 2^i\).

In turn, we can calculate the first \(6\) terms

• \(a_1 = 2^1 = 2\)
• \(a_2 = 2^2 = 4\) or calculated by \(a_2 = a_1 r = 2 \cdot 2 = 4\)
• \(a_3 = 2^3 = 8\) or calculated by \(a_3 = a_2 r = 4 \cdot 2 = 8\)
• \(a_4 = 2^4 = 16\) or calculated by \(a_4 = a_3 r = 8 \cdot 2 = 16\)
• \(a_5 = 2^5 = 32\) or calculated by \(a_5 = a_4 r = 16 \cdot 2 = 32\)
• \(a_6 = 2^6 = 64\) or calculated by \(a_6 = a_5 r = 32 \cdot 2 = 64\)

Check if the geometric series \(2 - \frac{4}{3} + \frac{8}{9} - \frac{16}{27} + \cdots\) converges or not. Find the sum if it converges.

This is a geometric series corresponding to the geometric sequence \(2, \frac{-4}{3}, \frac{8}{9}, \frac{-16}{27}, \dots\). The 1st term of the sequence is \(a_1 = 2\) and the 2nd term of the sequence is \(a_2 = \frac{-4}{3}\), so the common ratio of the sequence is \(r = \frac{a_2}{a_1} = \frac{\frac{-4}{3}}{2} = \frac{-2}{3} \in (-1,1)\).

Hence, by the rule, the series converges to \(\frac{a_1}{1-r} = \frac{2}{1-\frac{-2}{3}} = \frac{6}{5}\).

Check if the geometric series \(\sum_{i=1}^{\infty} \frac{5^{3i}}{2^{i-1}}\) converges or not. Find the sum if it converges.

This is a geometric series corresponding to the geometirc sequence whose general term is \(a_i = \frac{5^{3i}}{2^{i-1}}\). The 1st term of the sequence is \(a_1 = \frac{5^{3\cdot 1}}{2^{1-1}} = 125\) and the 2nd term of sequence is \(\frac{5^{3\cdot 2}}{2^{2-1}} = \frac{15625}{2}\), so the common ratio of the sequence is \(r = \frac{a_2}{a_1} = \frac{\frac{15625}{2}}{125} = \frac{125}{2} \notin (-1,1)\).

Hence, by the rule, the series diverges.

Express the repeated decimal \(0.535353\dots\) as a rational number.

A number is rational if it can be written as a fraction between two integers, so the nature of the problem is to convert \(0.535353\dots\) into a fraction of two integers.

We note that \(0.535353\dots = 0.53 + 0.0053 + 0.000053 + \cdots\) is a series corresponding to the sequence \(a_1 = 0.53, a_2 = 0.0053, a_3 = 0.000053, \dots\).

Now we verify that the sequence \(a_1 = 0.53, a_2 = 0.0053, a_3 = 0.000053, \dots\) is a geometric sequence : we see that \(\frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \cdots = \frac{a_{i+1}}{a_i} = 0.01\) is a constant for any \(i = 1, 2, 3, \dots\), so the sequence \(a_1 = 0.53, a_2 = 0.0053, a_3 = 0.000053, \dots\) is a geometric sequence with common ratio \(r = 0.01 \in (-1, 1)\).

Therefore, \(0.535353\dots = 0.53 + 0.0053 + 0.000053 + \cdots\) is the geometric series corresponding to the geometric sequence \(a_1 = 0.53, a_2 = 0.0053, a_3 = 0.000053, \dots\).

By the rule, sine \(r \in (-1,1)\), the geometric series \(0.535353\dots = 0.53 + 0.0053 + 0.000053 + \cdots\) converges to \(\frac{a_1}{1-r} = \frac{0.53}{1-0.01} = \frac{0.53}{0.99} = \frac{53}{99}\).

Finally, we have \(0.535353\dots = \frac{53}{99}\), which is a fraction between two integers, so it is a rational number.

Before applying L'Hospital's Rule, we should ALWAYS check whether the condition "\(\lim\limits_{x\to a} f(x) = 0\) and \(\lim\limits_{x\to a} g(x) = 0\) hold at the same time" or "\(\lim\limits_{x\to a} f(x) = \pm \infty\) and \(\lim\limits_{x\to a} g(x) = \pm\infty\) hold at the same time" holds. Otherwise, the rule does not apply.

Sometimes, after the first application of L'Hospital's Rule, the limit is still an indeterminate form. In this case, we may apply the L'Hospital's Rule for another time to see whether the problem gets simpler.

Find \(\lim\limits_{x\to 2} \frac{x^2 - 4}{x^2 - x - 6}\).

Here we identify

• \(f(x) = x^2 - 4\)
• \(g(x) = x^2 - x -6\)

We know

• \(\lim\limits_{x\to 2} f(x) = \lim\limits_{x\to 2} x^2 - 4 = 2^2 - 4 = 0\)
• \(\lim\limits_{x\to 2} g(x) = \lim\limits_{x\to 2} x^2 - x - 6 = 2^2 - 2 - 6 = -4 \neq 0\)

so the L'Hospital's Rule doesn't work for this problem.

As we learned in the Limit chapter, this limit can be found just by substitution:

\(\lim\limits_{x\to 2} \frac{x^2 - 4}{x^2 - x - 6} = \frac{0}{-4} = 0\).

Find \(\lim\limits_{x\to 2} \frac{x^2 - 4}{x^2 + x - 6}\).

Here we identify

• \(f(x) = x^2 - 4\)
• \(g(x) = x^2 + x -6\)

We know

• \(\lim\limits_{x\to 2} f(x) = \lim\limits_{x\to 2} x^2 - 4 = 2^2 - 4 = 0\)
• \(\lim\limits_{x\to 2} g(x) = \lim\limits_{x\to 2} x^2 + x - 6 = 2^2 + 2 - 6 = 0\)

so the limit is an indeterminate form.

By L'Hospital's Rule,

\(\lim\limits_{x\to 2} \frac{f(x)}{g(x)}\) \(= \lim\limits_{x\to 2} \frac{f'(x)}{g'(x)}\) \(= \lim\limits_{x\to 2} \frac{2x}{2x+1}\) \(= \frac{2 \cdot 2}{2\cdot 2 + 1}\) \(= \frac{4}{5}\)

Find \(\lim\limits_{x\to \infty} \frac{x^2 - 4}{x^2 - x - 6}\).

Here we identify

• \(f(x) = x^2 - 4\)
• \(g(x) = x^2 - x -6\)

We know

• \(\lim\limits_{x\to \infty} f(x)\) \(= \lim\limits_{x\to \infty} x^2 - 4\) \(= \infty^2 - 4 = \infty\)

• \(\lim\limits_{x\to \infty} g(x)\) \(= \lim\limits_{x\to \infty} x^2 - x - 6\) \(= \lim\limits_{x\to \infty} x (x - 1 - \frac{6}{x})\) \(= \lim\limits_{x\to \infty} x \cdot \lim\limits_{x\to \infty}x - 1 - \frac{6}{x}\) \(= \infty \cdot (\infty - 1 - 0)\) \(= \infty \cdot \infty = \infty\).

  (Note : for the denominator, if we directly substitute \(\infty\) for \(x\) in \(x^2 - x - 6\), then we get \(\infty^2 - \infty - 6 = \infty - \infty - 6\) and we don't know how to evaluate the indeterminate difference \(\infty - \infty\). We don't know whether it exists or not, and we don't know whether it is \(0\), \(\infty\) or any other constant. This is the reason why we do the factorization \(x^2 - x - 6 = x (x - 1 - \frac{6}{x})\) and then find \(\lim\limits_{x\to \infty} x^2 - x - 6 = \infty\).)

The limit is an indeterminate form \(\infty/\infty\).

By L'Hospital's Rule,

\(\lim\limits_{x\to \infty} \frac{f(x)}{g(x)}\) \(= \lim\limits_{x\to \infty} \frac{f'(x)}{g'(x)}\) \(= \lim\limits_{x\to \infty} \frac{2x}{2x+1}\) \(= \frac{2 \cdot \infty}{2\cdot \infty + 1}\) \(= \frac{\infty}{\infty}\)

which is again an indeterminate form \(\infty/\infty\).

Therefore, let's apply the L'Hospital's Rule for another time and get

\(\lim\limits_{x\to \infty} \frac{f(x)}{g(x)}\) \(= \lim\limits_{x\to \infty} \frac{f'(x)}{g'(x)}\) \(= \lim\limits_{x\to \infty} \frac{2x}{2x+1}\) \(= \lim\limits_{x\to \infty} \frac{(2x)'}{(2x+1)'}\) \(= \lim\limits_{x\to \infty} \frac{2}{2}\) \(= \lim\limits_{x\to \infty} 1\) \(= 1\).

Recall that we actually could find this limit without using L'Hospital's Rule. We can just divide both numerator and denominator by \(x^2\) and get:

\(\lim\limits_{x\to \infty} \frac{x^2 - 4}{x^2 - x - 6}\) \(= \lim\limits_{x\to \infty} \frac{\frac{x^2 - 4}{x^2}}{\frac{x^2 - x - 6}{x^2}}\) \(= \lim\limits_{x\to \infty} \frac{1 - \frac{4}{x^2}}{1 - \frac{1}{x} - \frac{6}{x^2}}\) \(= \frac{1 - \frac{4}{\infty^2}}{1 - \frac{1}{\infty} - \frac{6}{\infty^2}}\) \(= \frac{1 - 0}{1 - 0 - 0}\) \(= 1\).

Find \(\lim\limits_{x\to \infty} \frac{\sqrt{x+1}}{\sqrt{x-1}}\).

Here we identify

• \(f(x) = \sqrt{x+1}\)
• \(g(x) = \sqrt{x-1}\)

We know

• \(\lim\limits_{x\to \infty} f(x) = \lim\limits_{x\to \infty} \sqrt{x+1} = \sqrt{\infty+1} = \infty\)
• \(\lim\limits_{x\to \infty} g(x) = \lim\limits_{x\to \infty} \sqrt{x-1} = \sqrt{\infty-1} = \infty\)

The limit is an indeterminate form \(\infty/\infty\).

By L'Hospital's Rule,

\(\lim\limits_{x\to \infty} \frac{f(x)}{g(x)}\) \(= \lim\limits_{x\to \infty} \frac{f'(x)}{g'(x)}\) \(= \lim\limits_{x\to \infty} \frac{\frac{1}{2\sqrt{x+1}}}{\frac{1}{2\sqrt{x-1}}}\) \(= \lim\limits_{x\to \infty} \frac{1}{2\sqrt{x+1}} \cdot 2\sqrt{x-1} = \lim\limits_{x\to \infty} \frac{\sqrt{x-1}}{\sqrt{x+1}}\)

which is again an indeterminate form \(\infty/\infty\).

Should we apply the L'Hospital's Rule for one more time? Let's try it:

\(\lim\limits_{x\to \infty} \frac{\sqrt{x-1}}{\sqrt{x+1}}\) \(= \lim\limits_{x\to \infty} \frac{\frac{1}{2\sqrt{x-1}}}{\frac{1}{2\sqrt{x+1}}}\) \(= \lim\limits_{x\to \infty} \frac{1}{2\sqrt{x-1}} \cdot 2\sqrt{x+1} = \lim\limits_{x\to \infty} \frac{\sqrt{x+1}}{\sqrt{x-1}}\)

which is exactly the original problem!

As you can imagine, even if we apply the L'Hospital's Rule for even more times, the problem still doesn't get simpler.

Therefore, although L'Hospital's Rule is applicable here, it doesn't directly help us solve the problem. In this case, we need first make some change to the original problem.

We put the quantities under square root symbols together by

\(\lim\limits_{x\to \infty} \frac{\sqrt{x+1}}{\sqrt{x-1}}\) \(=\lim\limits_{x\to \infty} \sqrt{\frac{x+1}{x-1}}\) \(=\sqrt{\lim\limits_{x\to \infty} \frac{x+1}{x-1}}\)

Now we can find \(\lim\limits_{x\to \infty} \frac{x+1}{x-1}\) by

• L'Hospital's Rule, ie, \(\lim\limits_{x\to \infty} \frac{x+1}{x-1} = \lim\limits_{x\to \infty} \frac{1}{1} = 1\), or
• dividing both numerator and demoninator by \(x\), ie, \(\lim\limits_{x\to \infty} \frac{x+1}{x-1}\) \(= \lim\limits_{x\to \infty} \frac{\frac{x+1}{x}} {\frac{x-1}{x}}\) \(= \lim\limits_{x\to \infty} \frac{1 + \frac{1}{x}} {1 - \frac{1}{x}}\) \(= \frac{1 + 0} {1 - 0}\) \(= 1\).

Finally, the original limit

\(\lim\limits_{x\to \infty} \frac{\sqrt{x+1}}{\sqrt{x-1}}\) \(=\sqrt{\lim\limits_{x\to \infty} \frac{x+1}{x-1}}\) \(=\sqrt{\lim\limits_{x\to \infty} 1}\) \(=1\).