Week 9 Activity

Section 9.5 Week 9 Activity

Subsection 9.5.1 Sequences

Activity 9.5.1.

Find the limit of this sequence.

\begin{equation*} \lim_{n \rightarrow \infty} \frac{n^2}{n^3 + 3n +4} \end{equation*}

Solution.

I can use asymptotical analysis for this limit. The numerator has asymptotic order \(n^2\) and the denominator has asymptotic order \(n^3\text{.}\) The denominator has a higher asymptotic order, so the limit must be 0.

Activity 9.5.2.

Find the limit of this sequence.

\begin{equation*} \lim_{n \rightarrow \infty} \frac{2^n}{n!} \end{equation*}

Solution.

I can use asymptotical analysis for this limit, even though the factorial is a new expression. As mentioned in the notes, the factorial grows faster than exponential functions. Therefore, the denominator has a higher asymptotic order and the limit must be 0.

Activity 9.5.3.

Find the limit of this sequence.

\begin{equation*} \lim_{n \rightarrow \infty} \frac{n^2 + 3n}{n^2 + \sin (\pi n)} \end{equation*}

Solution.

I can use asymptotical analysis for this limit. The asymptotic order of the numerator is \(n^2\text{.}\) The asymptotic order of the denominator is the same (the sine function is bounded, so it has no effect on growth). The asymptotic orders are the same, so I look at the leading coefficients, which are \(\frac{1}{1}\text{.}\) Therefore, the limit is 1.

Activity 9.5.4.

Find the limit of this sequence.

\begin{equation*} \lim_{n \rightarrow \infty} \frac{\ln n}{n} \end{equation*}

Solution.

I can use asymptotical analysis for this limit. The logarithm grows slower than a linear function, so the denominator has a higher asymptotic order and the limit must be 0.

Activity 9.5.5.

Find the limit of this sequence.

\begin{equation*} \lim_{n \rightarrow \infty} (-1)^n \sin (\pi n) \end{equation*}

Solution.

This is a bit of a trick question. It looks like there are alternating terms, so like the sequences with terms \((-1)^n\text{,}\) there could be an argument that the limit doesn’t exist. However, every term in the limit has \(\sin (\pi n)\) and, for sequences, \(n \in \NN\text{.}\) Sine of any multiple of \(\pi\) is zero, so all the terms of the sequence are 0. Therefore, the limit is also zero.

Activity 9.5.6.

Find the limit of this sequence.

\begin{equation*} \lim_{n \rightarrow \infty} \frac{(-1)^n}{\sqrt{n}} \end{equation*}

Solution.

The denoiminator is growing arbitrarity large, since the square root is an increasing, unbounded function. Dividing by large numbers give small numbers, so the ration is approaching zero. The fact that it keeps changing sign between positive and negative doesn’t matter, since the terms are still getting closer and closer to zero. The limit is 0.

Activity 9.5.7.

Find the limit of this sequence.

\begin{equation*} \lim_{n \rightarrow \infty} \tan (n) \end{equation*}

Solution.

This is an extremely chaotic limit. The function \(\tan n\) take all kinds of values since the argument is not a multiple of \(\pi\) and no regular behaviour can be expected. Since the terms are just jumping around all over the place, there is no convergence and the limit cannot exist.

Activity 9.5.8.

Find the limit of this sequence.

\begin{equation*} \lim_{ n \rightarrow \infty} \frac{e^n}{e^{2n} + 1} \end{equation*}

Solution.

I can use asymptotical analysis for this limit. The asymptotic order of the numerator is \(e^n\text{.}\) The asymptotic order of the denominator is \(e^{2n} = (e^n)^2 = e^n e^n \text{.}\) The denominator has a higher asymptotic order, so the limit must be zero.

Activity 9.5.9.

Assume that \(a_n\) are the terms of a recursive sequnce with \(a_1 = 1\) and with the following recursive relationship.

\begin{equation*} a_n = 3 + \frac{2}{a_{n-1}} \end{equation*}

What is the limit of the sequence?

Solution.

I can take the limit of both sides of the recursion equation. The limit of \(a_n\) or of \(a_{n-1}\) is still the limit of the sequence. I’ll call that limit \(L\) and solve for it using the equation.

\begin{gather*} \lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} 3 + \frac{2}{a_{n-1}} \\ \lim_{n \rightarrow \infty} a_n = 3 + \frac{2}{\lim_{n \rightarrow \infty} a_{n-1}} \\ L = 3 + \frac{2}{L} \\ L^2 = 3L + 2\\ L^2 - 3L - 2 = 0 \\ L = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2} \end{gather*}

This is interest, there are two possible limit. Which one is the limit of this sequence? Well, one limit is positive and one limit is negative. The sequences starts with \(a_1 = 1\) and the recursion relation only produces positive numbers, so the limit must be positive. I conclude the limit is \(L = \frac{3 + \sqrt{17}}{2}\text{.}\)

Activity 9.5.10.

Assume that \(a_n\) are the terms of a recursive sequnce with \(a_1 = 1\) and with the following recursive relationship.

\begin{equation*} a_n = 1 - \frac{1}{a_{n-1}} \end{equation*}

What is the limit of the sequence?

Solution.

\begin{gather*} \lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} 1 - \frac{1}{a_{n-1}} \\ \lim_{n \rightarrow \infty} a_n = 1 - \frac{1}{\lim_{n \rightarrow \infty} a_{n-1}} \\ L = 1 - \frac{1}{L} \\ L^2 = L - 1\\ L^2 - L + 1 = 0 \\ L = \frac{1 + \sqrt{-3}}{2} \end{gather*}

The quadratic has no solutions. Does that mean there is no limit? Indeed, it does. Even more, this sequence has more significant problems. With \(a_1 = 1\text{,}\) the recusion relation gives \(a_2 = 0\text{.}\) Then if I try to calcualte \(a_3\text{,}\) there is division by zero. Therefore, this sequences isn’t even defined, so it certainly can’t have a limit.

Activity 9.5.11.

This question generalizes the two previous questions, trying to look for a general statement about limits of this type of recursive sequence. Assume that \(a_n\) are the terms of a recursive sequnce with \(a_1 = 1\) and with the following recursive relationship, where \(\alpha, \beta \in \RR\text{.}\)

What is the limit of the sequence? How does the limit depend on \(\alpha\) and \(\beta\text{?}\) What are the conditions on \(\alpha\) and \(\beta\) that allow for the limit to exists? What are the possible behaviours?

Solution.

I use the same starting setup as the previous limit.

\begin{gather*} \lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \alpha + \frac{\beta}{a_{n-1}} \\ \lim_{n \rightarrow \infty} a_n = \alpha+ \frac{\beta}{\lim_{n \rightarrow \infty} a_{n-1}} \\ L = \alpha + \frac{\beta}{L} \\ L^2 = \alpha L + \beta\\ L^2 - \alpha L - \beta = 0 \\ L = \frac{\alpha \pm \sqrt{ \alpha^2 + 4\beta}}{2} \end{gather*}

The detemination of ths limit depends on the roots of this quadratic. If the quadratic has no roots, if \(\alpha^2 + 4 \beta \lt 0\text{,}\) then the sequence will not converge to any limit. If \(\alpha^2 + 4 \beta = 0\text{,}\) then there is a unique root of \(\frac{\alpha}{2}\) and the sequence will approach that root. Finally, if \(\alpha^2 + 4 \beta \gt 0\text{,}\) theen there are two roots and the sequence will appraoch one of the two roots depending on the starting value.

In addition to this discussion, I also have to be aware that some starting values will lead to division by zero in the recurrence relation, as happened in the previous question. That is another way in which the sequence can fail to have a limit.

Subsection 9.5.2 Convergence of Series

Activity 9.5.12.

Use the limit of partial sum to calculate the value of this infinite series. (Hint: multiply by \(\frac{r-1}{r-1}\) for some value of \(r\) to simplify the expression for the partial sum.)

\begin{equation*} \sum_{n=1}^\infty \frac{1}{3^n} \end{equation*}

Solution.

I start by calculating the first few parial sums. I try to produce a pattern from these partial sums.

\begin{align*} s_1 \amp = \frac{1}{3} \\ s_2 \amp = \frac{1}{3} + \frac{1}{9} = \frac{3+1}{9} = \frac{4}{9} \\ s_3 \amp = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} = \frac{9+3+1}{27} = \frac{13}{27} \\ s_4 \amp = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} = \frac{3^3 + 3^2 + 3 + 1}{81} \\ \amp = \frac{(3-1)(3^3 + 3^2 + 3 + 1)}{3^4(3-1)} = \frac{3^4 - 1}{3^4 (3-1)} \end{align*}

Multiplying by \(\frac{(3-1)}{(3-1)}\) and distributing out the numerator lets most of the terms cancel off, leading to a more reasoanble expression and a general pattern.

\begin{align*} s_5 \amp = \frac{3^5 - 1}{3^5(2)} \\ s_6 \amp = \frac{3^6 - 1}{3^6(2)} \\ s_n \amp = \frac{3^n - 1}{3^n(2)} \end{align*}

The value of the series is the limit of the partial sums. Here that limit is \(\frac{1}{2}\text{.}\)

Activity 9.5.13.

Use the limit of partial sum to calculate the value of this infinite series. (Hint: multiply by \(\frac{r-1}{r-1}\) for some value of \(r\) to simplify the expression for the partial sum.)

\begin{equation*} \sum_{n=1}^\infty \frac{1}{5^n} \end{equation*}

Solution.

I start by calculating the first few parial sums. I try to produce a pattern from these partial sums.

\begin{align*} s_1 \amp = \frac{1}{5} \\ s_2 \amp = \frac{1}{5} + \frac{1}{25} = \frac{4+1}{25} = \frac{5}{25}\\ s_3 \amp = \frac{1}{5} + \frac{1}{25} + \frac{1}{125} = \frac{25+5+1}{125} = \frac{31}{125} \\ s_4 \amp = \frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \frac{1}{625} = \frac{5^3 + 5^2 + 5 + 1}{625} \\ \amp = \frac{(5-1)(5^3 + 5^2 + 5 + 1)}{5^4(5-1)} = \frac{5^4 - 1}{5^4 (5-1)} \end{align*}

Multiplying by \(\frac{(5-1)}{(5-1)}\) and distributing out the numerator lets most of the terms cancel off, leading to a more reasoanble expression and a general pattern.

\begin{align*} s_5 \amp = \frac{5^5 - 1}{5^5(4)} \\ s_6 \amp = \frac{5^6 - 1}{5^6(4)} \\ s_n \amp = \frac{5^n - 1}{5^n(4)} \end{align*}

The value of the series is the limit of the partial sums. Here that limit is \(\frac{1}{4}\text{.}\)

Activity 9.5.14.

Use the limit of partial sum to calculate the value of this infinite series. (Hint: multiply by \(\frac{r-1}{r-1}\) for some value of \(r\) to simplify the expression for the partial sum.)

\begin{equation*} \sum_{n=1}^\infty \frac{1}{10^n} \end{equation*}

Solution.

I start by calculating the first few parial sums. I try to produce a pattern from these partial sums.

\begin{align*} s_1 \amp = \frac{1}{10} \\ s_2 \amp = \frac{1}{10} + \frac{1}{100} = \frac{10+1}{100} = \frac{11}{100} \\ s_3 \amp = \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} = \frac{100+10+1}{1000} = \frac{111}{1000}\\ s_4 \amp = \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \frac{1}{10000} = \frac{(10-1)(10^3 + 10^2 + 10 + 1)}{10^4(10-1)} \\ \amp = \frac{10^4 - 1}{10^4(10-1)} \end{align*}

Multiplying by \(\frac{(10-1)}{(10-1)}\) and distributing out the numerator lets most of the terms cancel off, leading to a more reasoanble expression and a general pattern.

\begin{align*} s_5 \amp = \frac{10^5 - 1}{10^5(9)} \\ s_6 \amp = \frac{10^6 - 1}{10^6(9)} \\ s_n \amp = \frac{10^n - 1}{10^n(9)} \end{align*}

The value of the series is the limit of the partial sums. Here that limit is \(\frac{1}{9}\text{.}\)

Activity 9.5.15.

Use the limit of partial sum to calculate the value of this infinite series. (Hint: use partial fracitons to expand the term into two fractions. Then, in the partial terms, cancel most of the fractions.)

\begin{equation*} \sum_{n=1}^\infty \frac{1}{n^2 + 4n + 3} \end{equation*}

Solution.

First, I do partial fractions for the rational function. I’ve not shown the steps of partial fraction in this key, just the result.

\begin{align*} \frac{1}{n^2 + 4n + 3} \amp = \frac{1}{(n+3)(n+1)} = \frac{\frac{1}{2}}{n+1} + \frac{\frac{-1}{2}}{n+3} \\ \sum_{n=1}^\infty \frac{1}{n^2 + 3n + 4} \amp = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n+1} - \frac{1}{n+3} \end{align*}

The I calculate the first few parial sums. I try to produce a pattern from these partial sums.

\begin{align*} s_1 \amp = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{4} \right] \\ s_2 \amp = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} \right]\\ s_3 \amp = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6}\right] \\ \amp = \frac{1}{2} \left[ \frac{1}{2} + \frac{1}{3} - \frac{1}{5} - \frac{1}{6}\right] \\ s_4 \amp = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \frac{1}{5} - \frac{1}{7} \right]\\ \amp = \frac{1}{2} \left[ \frac{1}{2} + \frac{1}{3} - \frac{1}{6} - \frac{1}{7}\right]\\ s_5 \amp = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \frac{1}{5} - \frac{1}{7} + \frac{1}{6} - \frac{1}{8} \right]\\ \amp = \frac{1}{2} \left[ \frac{1}{2} + \frac{1}{3} - \frac{1}{7} - \frac{1}{8}\right] \end{align*}

This is enough for me to conclude a pattern.

\begin{align*} s_n \amp = \frac{1}{2} \left[ \frac{1}{2} + \frac{1}{3} - \frac{1}{n+2} - \frac{1}{n+3} \right] \end{align*}

The value of the series is the limit of the partial sums.

\begin{align*} \sum_{n=1}^\infty \frac{1}{n^2 + 3n + 4} \amp = \lim_{n \rightarrow \infty} \frac{1}{2} \left[ \frac{1}{2} + \frac{1}{3} - \frac{1}{n+2} - \frac{1}{n+3} \right] \\ \amp = \frac{1}{2} \left[ \frac{1}{2} + \frac{1}{3} \right] = \frac{5}{12} \end{align*}

Activity 9.5.16.

\begin{equation*} \sum_{n=1}^\infty \frac{1}{n^2 + 5n + 6} \end{equation*}

Solution.

First, I do partial fractions for the rational function. I’ve not shown the steps of partial fraction in this key, just the result.

\begin{align*} \frac{1}{n^2 + 5n + 6} \amp = \frac{1}{(n+3)(n+2)} = \frac{1}{n+2} + \frac{-1}{n+3}\\ \sum_{n=1}^\infty \frac{1}{n^2 + 5n + 6} \amp = \sum_{n=1}^\infty \frac{1}{n+2} - \frac{1}{n+3} \end{align*}

The I calculate the first few parial sums. I try to produce a pattern from these partial sums.

\begin{align*} s_1 \amp = \frac{1}{3} - \frac{1}{4} \\ s_2 \amp = \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5}\\ s_3 \amp = \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} = \frac{1}{3} - \frac{1}{6}\\ s_4 \amp = \frac{1}{3} - \frac{1}{7} \\ s_5 \amp = \frac{1}{3} - \frac{1}{8} \end{align*}

This is enough for me to conclude a pattern.

\begin{align*} s_n \amp = \frac{1}{3} - \frac{1}{n+3} \end{align*}

The value of the series is the limit of the partial sums.

\begin{align*} \sum_{n=1}^\infty \frac{1}{n^2 + 3n + 4} \amp = \lim_{n \rightarrow \infty} \frac{1}{3} - \frac{1}{n+3} = \frac{1}{3} \end{align*}

Activity 9.5.17.

Consider this series.

\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{3n} \end{equation*}

Why does this series diverge? (You can argue directly or look at the pattern of partial sums to make an argument that the limit of partial sums cannot converge.)

Solution.

This series diverges because it is \(\frac{1}{3}\) multiplied by the harmonic series. We know the harmonic series diverges, and \(\frac{1}{3}\) of something growing to \(\infty\) still grows to \(\infty\text{.}\)

Activity 9.5.18.

Consider this series.

\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \end{equation*}

Why does this series diverge? (You can argue directly or look at the pattern of partial sums to make an argument that the limit of partial sums cannot converge.)

Solution.

The terms \(\frac{1}{\sqrt{n}}\) are larger than the terms of the harmonic series, to this series also diverges.

Activity 9.5.19.

Consider this series.

\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{n^2 - 9n + 20} \end{equation*}

Why does this series diverge? (You can argue directly or look at the pattern of partial sums to make an argument that the limit of partial sums cannot converge.)

Solution.

I look at the first few partial sums

\begin{align*} s_1 \amp = \frac{1}{1-9+20} = \frac{1}{12} \\ s_2 \amp = \frac{1}{12} + \frac{1}{4-18+20} = \frac{1}{12} + \frac{1}{6} = \frac{1}{4}\\ s_3 \amp = \frac{1}{4} + \frac{1}{9 - 27 + 20} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\\ s_4 \amp = \frac{3}{4} + \frac{1}{16-36+20} \end{align*}

The fourth partial sum is undefined, since the denominator evaluates to zero. This sum cannot converge, since its terms are not well defined.

Activity 9.5.20.

Consider this series.

\begin{equation*} \sum_{n=1}^{\infty} (-1)^{\frac{n(n+1)}{2}} \end{equation*}

Why does this series diverge? (You can argue directly or look at the pattern of partial sums to make an argument that the limit of partial sums cannot converge.)

Solution.

The test for divergence shows that this cannot converge. The terms are either \(1\) or \(-1\text{,}\) so the limit of the terms cannot approach \(0\text{.}\) In the partial sums, we are always either adding or subtracting \(1\text{,}\) which will never converge to any value.

Activity 9.5.21.

Consider this series.

\begin{equation*} \sum_{n=1}^{\infty} \sin \left( \frac{n\pi}{2} \right) \end{equation*}

Why does this series diverge? (You can argue directly or look at the pattern of partial sums to make an argument that the limit of partial sums cannot converge.)

Solution.

I look at the first few partial sums.

\begin{align*} s_1 \amp = \sin \left( \frac{\pi}{2} \right) = 1 \\ s_2 \amp = 1 + \sin \left( \frac{2\pi}{2} \right) = 1 + 0 = 1 \\ s_3 \amp = 1 + \sin \left( \frac{3\pi}{2} \right) = 1 + -1 = 0 \\ s_4 \amp = 0 + \sin \left( \frac{4\pi}{2} \right) = 0 + 0 = 0 \\ s_5 \amp = 0 + \sin \left( \frac{5\pi}{2} \right) = 0 + 1 = 1 \end{align*}

These partial sums will endlessly oscillate between \(0\) and \(1\text{.}\) There is no limit to the sequence of partial sums, so the series cannot converge.

Activity 9.5.22.

Consider this series.

\begin{equation*} \sum_{n=1}^{\infty} 2^n \end{equation*}

Why does this series diverge? (You can argue directly or look at the pattern of partial sums to make an argument that the limit of partial sums cannot converge.)

Solution.

I look at the first few partial sums.

\begin{align*} s_1 \amp = 2 \\ s_2 \amp = 2 + 4 = 6 \\ s_3 \amp = 2 + 4 + 8 = 14 \\ s_4 \amp = 2 + 4 + 8 + 16 = 30 \\ s_5 \amp = 2 + 4 + 8 + 16 + 32 = 62 \end{align*}

These partial sums are simply unbouded. The series grows to infinity. Equivalently, the test for divergence and the fact that the terms do not go to zero shows that convergence is impossible.

Activity 9.5.23.

Consider this series.

\begin{equation*} \sum_{n=1}^{\infty} \ln \left( \frac{1}{n} \right) \end{equation*}

Why does this series diverge? (You can argue directly or look at the pattern of partial sums to make an argument that the limit of partial sums cannot converge.)

Solution.

I look at the first few partial sums.

\begin{align*} s_1 \amp = \ln 1 = 0 \\ s_2 \amp = 0 + \ln \frac{1}{2} \\ s_3 \amp = \ln \frac{1}{2} + \ln \frac{1}{3} = \ln \frac{1}{6} \\ s_4 \amp = \ln \frac{1}{6} + \ln \frac{1}{4} = \ln \frac{1}{24} = \ln \frac{1}{4!} \\ s_n \amp = \ln \frac{1}{n!} \end{align*}

As \(\frac{1}{n!}\) tends towards zero, the value of the logarithm tends towards \(-\infty\text{,}\) so this series will also diverge to \(-\infty\text{.}\)

Subsection 9.5.3 Conceptual Review Questions

What is a sequence? How is it like a function?
How do you calculate the limit of a sequence? How is this different from limits of functions?
What is an infinite series? What does it mean for it to converge?
What is a geometric series? What is a zeta series?

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