Infinite Sequences
Solved Problems
Example 3.
Determine whether the sequence \[\left\{ {\frac{{2n + 3}}{{5n - 7}}} \right\}\] converges or diverges.
Solution.
Divide by the highest power in the numerator and denominator:
\[\lim\limits_{n \to \infty } \frac{{2n + 3}}{{5n - 7}} = \lim\limits_{n \to \infty } \frac{{\frac{{2n + 3}}{n}}}{{\frac{{5n - 7}}{n}}} = \lim\limits_{n \to \infty } \frac{{2 + \frac{3}{n}}}{{5 - \frac{7}{n}}} = \frac{2}{5}.\]
Hence, the sequence converges to \(\frac{2}{5}.\)
Example 4.
Does the sequence \[\left\{ {\frac{{{n^2}}}{{{2^n}}}} \right\}\] converge or diverge?
Solution.
As L'Hopital's rule yields
\[\lim\limits_{x \to \infty } \left( { \pm \frac{x}{{{2^{x - 1}}}}} \right) = \pm \lim\limits_{x \to \infty } \frac{x}{{{2^{x - 1}}}} = \pm \lim\limits_{x \to \infty } \frac{1}{{{2^{x - 1}}\ln 2}} = 0.\]
Since the limit is finite, the given sequence converges.
Example 5.
Determine whether the sequence \[\left\{ {\sqrt {n + 2} - \sqrt {n + 1} } \right\}\] converges or diverges.
Solution.
We multiply this expression by the quotient
\[\frac{{\sqrt {n + 2} + \sqrt {n + 1} }}{{\sqrt {n + 2} + \sqrt {n + 1} }} = 1\]
to obtain:
\[\lim\limits_{n \to \infty } \left( {\sqrt {n + 2} - \sqrt {n + 1} } \right) = \lim\limits_{n \to \infty } \left( {\sqrt {n + 2} - \sqrt {n + 1} } \right)\cdot \frac{{\sqrt {n + 2} + \sqrt {n + 1} }}{{\sqrt {n + 2} + \sqrt {n + 1} }} = \lim\limits_{n \to \infty } \frac{{{{\left( {\sqrt {n + 2} } \right)}^2} - {{\left( {\sqrt {n + 1} } \right)}^2}}}{{\sqrt {n + 2} + \sqrt {n + 1} }} = \lim\limits_{n \to \infty } \frac{{n + 2 - \left( {n + 1} \right)}}{{\sqrt {n + 2} + \sqrt {n + 1} }} = \lim\limits_{n \to \infty } \frac{1}{{\sqrt {n + 2} + \sqrt {n + 1} }} = 0.\]
This means that the sequence converges.
Example 6.
Determine whether the sequence \[\left\{ {\frac{{5n - 7}}{{3n + 4}}} \right\}\] is increasing, decreasing, or neither.
Solution.
The \(\left( {n + 1} \right)\)th term of the sequence is given by the formula
\[a_{n + 1} = \frac{{5\left( {n + 1} \right) - 7}}{{3\left( {n + 1} \right) + 4}} = \frac{{5n - 2}}{{3n + 7}}.\]
Check the inequality \({a_n} \le {a_{n + 1}}:\)
\[\frac{{5n - 7}}{{3n + 4}} \le \frac{{5n - 2}}{{3n + 7}},\;\; \Rightarrow \frac{{5n - 7}}{{3n + 4}} - \frac{{5n - 2}}{{3n + 7}} \le 0,\;\; \Rightarrow \frac{{ - {41}}}{{\left( {3n + 4} \right)\left( {3n + 7} \right)}} \le 0.\]
The last inequality is obvious, since the numerator is negative and \(3n + 4 \ge 0\) and \(3n + 7 \ge 0\) for \(n \ge 1.\) Therefore, this sequence is increasing.
Example 7.
Determine whether the sequence \[\left\{ {\frac{{{2^n} + 3}}{{{2^n} + 1}}} \right\}\] is increasing, decreasing, or not monotonic.
Solution.
Write out the first few terms of the sequence:
\[\left\{ {\frac{{{2^n} + 3}}{{{2^n} + 1}}} \right\} = \left\{ {\frac{5}{3},\frac{7}{5},\frac{{11}}{9},\frac{{19}}{{17}},\frac{{35}}{{33}}, \ldots } \right\}.\]
We see that this is a decreasing sequence. To confirm this, we prove the inequality \({a_n} \ge {a_{n + 1}}.\) We have
\[{a_n} = \frac{{{2^n} + 3}}{{{2^n} + 1}},\;\; {a_{n + 1}} = \frac{{{2^{n + 1}} + 3}}{{{2^{n + 1}} + 1}}.\]
Then the condition \({a_n} \ge {a_{n + 1}}\) implies that
\[\frac{{{2^n} + 3}}{{{2^n} + 1}} \ge \frac{{{2^{n + 1}} + 3}}{{{2^{n + 1}} + 1}}.\]
Multiply both sides of the inequality by \(\left( {{2^n} + 1} \right)\) \(\left( {{2^{n + 1}} + 1} \right):\)
\[\left( {{2^n} + 3} \right)\left( {{2^{n + 1}} + 1} \right) \ge \left( {{2^{n + 1}} + 3} \right)\left( {{2^n} + 1} \right),\;\; \Rightarrow {2^n}{2^{n + 1}} + 3 \cdot {2^{n + 1}} + {2^n} + 3 \ge {2^n}{2^{n + 1}} + 3 \cdot {2^n} + {2^{n + 1}} + 3,\;\; \Rightarrow 2 \cdot {2^{n + 1}} \ge 2 \cdot {2^n},\;\; \Rightarrow {2^{n + 1}} \ge {2^n},\;\;\Rightarrow 2 \ge 1.\]
Since the last inequality is true, we can conclude that the sequence is decreasing.
