Tag: Sequences

Prove that a sequence cannot converge to two different limits

by RoRi

February 29, 2016

Prove that a sequence cannot converge to two different limits.

Proof. Consider a sequence $\{ a_n \}$ such that

$\lim_{n \to \infty} a_n = L \qquad \text{and} \qquad \lim_{n \to \infty} a_n = L^*.$

We show that we must have $L = L^*$ . By the definition of the limit we know that for all $\varepsilon > 0$ there exist a positive integers $N_1$ and $N_2$ such that

$\begin{align*} |a_n - L| &< \frac{\varepsilon}{2} & \text{for all } n \geq N_1 \\ |a_n - L^*| &< \frac{\varepsilon}{2} & \text{for all } n \geq N_2. \end{align*}$

Let $N = \max \{ N_1, N_2 \}$ , then for all $n \geq N$ and all $\varepsilon > 0$ we have

$|a_n - L| < \frac{\varepsilon}{2} \qquad \text{and} \qquad |a_n - L^*| < \frac{\varepsilon}{2}.$

Hence,

$\begin{align*} &&|a_n - L| + |L^* - a_n| &< \varepsilon \\ \implies && |a_n - L + L^* - a_n| &< \varepsilon \\ \implies && |L^* - L| &< \varepsilon \\ \implies && L = L^*. \end{align*}$

(The final line follows since if $L \neq L^*$ , then $|L^* - L| > 0$ , so setting $\varepsilon = |L - L^*|$ would contradict that $|L^* - L| < \varepsilon$ for all $\varepsilon > 0$ .) Therefore, a sequence cannot converge to two different limits $. \qquad \blacksquare$

Find an N such that the given convergent sequence is within ε of its limit

by RoRi

February 29, 2016

Consider the convergent sequence $\{ a_n \}$ with terms defined by

$a_n = (-1)^n \left( \frac{9}{10} \right)^n.$

Let $L = \lim_{n \to \infty} a_n$ . Find the value of $L$ and the value of $N$ such that $|a_n - L| < \varepsilon$ for all $n \geq N$ for each of the following values of $\varepsilon$ :

$\varepsilon = 1$ ,
$\varepsilon = 0.1$ ,
$\varepsilon = 0.01$ ,
$\varepsilon = 0.001$ ,
$\varepsilon = 0.0001$ .

First, we know

$\lim_{n \to \infty} (-1)^n \left( \frac{9}{10}\right)^n = \lim_{n \to \infty} \left(-\frac{9}{10}\right)^n = 0$

since $\left|\frac{-9}{10}\right| = \frac{9}{10} < 1$ , so by (10.10) on page 380 of Apostol we know the limit is 0.
Then we have,

$\begin{align*} |a_n - L| < \varepsilon && \implies && \left| \left(-\frac{9}{10} \right)^n \right| &< \varepsilon \\[9pt] && \implies && \left( \frac{9}{10} \right)^n &< \varepsilon \\[9pt] && \implies && \log \left(\frac{9}{10}\right)^n &< \log \varepsilon \\[9pt] && \implies && n \log \frac{9}{10} &< \log \varepsilon \\[9pt] && \implies && n &> \frac{\log \varepsilon}{\log \frac{9}{10}}. \end{align*}$

We reversed the inequality sign in the final step since $\log \frac{9}{10} < 0$ since $\frac{9}{10} < 1$ . Thus, if $N > \frac{\log \varepsilon}{\log \frac{9}{10}}$ then for every $n \geq N$ we have $|a_n| < \varepsilon$ . We compute for the given values of $\varepsilon$ as follows:

$\varepsilon = 1$ implies $N = \frac{\log 1}{\log 0.9} = 0$ .
$\varepsilon = 0.1$ implies $N = \frac{\log 0.1}{\log 0.9}$ .
$\varepsilon = 0.01$ implies $N = \frac{\log 0.01}{\log 0.9}$ .
$\varepsilon = 0.001$ implies $N = \frac{\log 0.001}{\log 0.9}$ .
$\varepsilon = 0.0001$ implies $N = \frac{\log 0.0001}{\log 0.9}$ .

Find an N such that the given convergent sequence is within ε of its limit

by RoRi

February 28, 2016

Consider the convergent sequence $\{ a_n \}$ with terms defined by

$a_n = \frac{2n}{n^3 + 1}.$

Let $L = \lim_{n \to \infty} a_n$ . Find the value of $L$ and values of $N$ such that $|a_n - L| < \varepsilon$ for all $n \geq N$ for each of the following values of $\varepsilon$ :

$\varepsilon = 1$ ,
$\varepsilon = 0.1$ ,
$\varepsilon = 0.01$ ,
$\varepsilon = 0.001$ ,
$\varepsilon = 0.0001$ .

First, we know

$\lim_{n \to \infty} \frac{2n}{n^3 + 1} = \lim_{n \to \infty} \frac{\frac{2}{n^2}}{1 + \frac{1}{n^3}} = 0 \quad \implies \quad L= 0.$

So then we have,

$\begin{align*} |a_n - L| < \varepsilon && \implies && \left| \frac{2n}{n^3+1} \right| &< \varepsilon \\[9pt] && \implies && \frac{2n}{n^3 + 1} &< \varepsilon \\[9pt] && \implies && 2n &< \varepsilon (n^3+1) \\[9pt] && \implies && \frac{2}{\varepsilon} &< n^2 + \frac{1}{n}. \end{align*}$

Thus, if $N > \sqrt{\frac{2}{\varepsilon}}$ then for every $n \geq N$ we have $|a_n| < \varepsilon$ . We compute for the given values of $\varepsilon$ as follows:

$\varepsilon = 1$ implies $N \geq \sqrt{\frac{2}{1}} = \sqrt{2}$ .
$\varepsilon = 0.1$ implies $N \geq \sqrt{\frac{2}{.1}} = \sqrt{20} = 2 \sqrt{5}$ .
$\varepsilon = 0.01$ implies $N \geq \sqrt{\frac{2}{.01}} = \sqrt{200} = 10\sqrt{2}$ .
$\varepsilon = 0.001$ implies $N \geq \sqrt{\frac{2}{.001}} = \sqrt{2000} = 20 \sqrt{5}$ .
$\varepsilon = 0.0001$ implies $N \geq \sqrt{\frac{2}{.0001}} = \sqrt{20000} = 100 \sqrt{2}$ .

Find an N such that the convergent sequence is within ε of its limit

by RoRi

February 28, 2016

Consider the convergent sequence $\{ a_n \}$ with terms defined by

$a_n = \frac{(-1)^{n+1}}{n}.$

Let $L = \lim_{n \to \infty} a_n$ . Find the value of $L$ and values of $N$ such that $|a_n - L| < \varepsilon$ for all $n \geq N$ for each of the following values of $\varepsilon$ :

$\varepsilon = 1$ ,
$\varepsilon = 0.1$ ,
$\varepsilon = 0.01$ ,
$\varepsilon = 0.001$ ,
$\varepsilon = 0.0001$ .

First, we know

$\lim_{n \to \infty} \frac{(-1)^{n+1}}{n} \leq \lim_{n \to \infty} \left| \frac{(-1)^{n+1}}{n} \right| = \lim_{n \to \infty} \frac{1}{n} = 0 \quad \implies \quad L= 0.$

So then we have,

$\begin{align*} |a_n - L| < \varepsilon && \implies && \left| \frac{(-1)^{n+1}}{n} \right| &< \varepsilon \\ && \implies && \frac{1}{n} &< \varepsilon \\ && \implies && n &> \frac{1}{\varepsilon}. \end{align*}$

Thus, if $N > \frac{1}{\varepsilon}$ then for every $n \geq N$ we have $|a_n| < \varepsilon$ . We compute for the given values of $\varepsilon$ as follows:

$\varepsilon = 1$ implies $N = \frac{1}{1} = 1$ .
$\varepsilon = 0.1$ implies $N = \frac{1}{.1} = 10$ .
$\varepsilon = 0.01$ implies $N = \frac{1}{.01} = 100$ .
$\varepsilon = 0.001$ implies $N = \frac{1}{.001} = 1000$ .
$\varepsilon = 0.0001$ implies $N = \frac{1}{.0001} = 10000$ .

Find an N such that the convergent sequence is within ε of its limit

by RoRi

February 28, 2016

Consider the convergent sequence $\{ a_n \}$ with terms defined by

$a_n = \frac{n}{n+1}.$

Let $L = \lim_{n \to \infty} a_n$ . Find the value of $L$ and values of $N$ such that $|a_n - L| < \varepsilon$ for all $n \geq N$ for each of the following values of $\varepsilon$ :

$\varepsilon = 1$ ,
$\varepsilon = 0.1$ ,
$\varepsilon = 0.01$ ,
$\varepsilon = 0.001$ ,
$\varepsilon = 0.0001$ .

First, we know

$\lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1+ \frac{1}{n}} = 1 \quad \implies \quad L= 1.$

So then we have,

$\begin{align*} |a_n - L| < \varepsilon && \implies && \left| \frac{n}{n+1} - 1 \right| &< \varepsilon \\ && \implies && \frac{n-n-1}{n+1} &< \varepsilon \\ && \implies && \frac{1}{n+1} &< \varepsilon \\ && \implies && \frac{1}{\varepsilon} &< n+1. \end{align*}$

Thus, if $N > \frac{1}{\varepsilon}$ then for every $n \geq N$ we have $|a_n| < \varepsilon$ . We compute for the given values of $\varepsilon$ as follows:

$\varepsilon = 1$ implies $N = \frac{1}{1} = 1$ .
$\varepsilon = 0.1$ implies $N = \frac{1}{.1} = 10$ .
$\varepsilon = 0.01$ implies $N = \frac{1}{.01} = 100$ .
$\varepsilon = 0.001$ implies $N = \frac{1}{.001} = 1000$ .
$\varepsilon = 0.0001$ implies $N = \frac{1}{.0001} = 10000$ .

Find an N such that the convergent sequence is within ε of its limit

by RoRi

February 28, 2016

Consider the convergent sequence $\{ a_n \}$ with terms defined by

$a_n = \frac{1}{n}.$

Let $L = \lim_{n \to \infty} a_n$ . Find the value of $L$ and values of $N$ such that $|a_n - L| < \varepsilon$ for all $n \geq N$ for each of the following values of $\varepsilon$ :

$\varepsilon = 1$ ,
$\varepsilon = 0.1$ ,
$\varepsilon = 0.01$ ,
$\varepsilon = 0.001$ ,
$\varepsilon = 0.0001$ .

First, we know

$\lim_{n \to \infty} \frac{1}{n} = 0 \quad \implies \quad L= 0.$

So then we have,

$\begin{align*} |a_n - L| < \varepsilon && \implies && \left| \frac{1}{n} \right| &< \varepsilon \\ && \implies && \frac{1}{n} &< \varepsilon \\ && \implies && n &> \frac{1}{\varepsilon}. \end{align*}$

Thus, if $N > \frac{1}{\varepsilon}$ then for every $n \geq N$ we have $|a_n| < \varepsilon$ . We compute for the given values of $\varepsilon$ as follows:

$\varepsilon = 1$ implies $N = \frac{1}{1} = 1$ .
$\varepsilon = 0.1$ implies $N = \frac{1}{.1} = 10$ .
$\varepsilon = 0.01$ implies $N = \frac{1}{.01} = 100$ .
$\varepsilon = 0.001$ implies $N = \frac{1}{.001} = 1000$ .
$\varepsilon = 0.0001$ implies $N = \frac{1}{.0001} = 10000$ .

Determine the convergence or divergence of f(n) = ne^{-πin / 2}

by RoRi

February 28, 2016

Consider the function $f(n)$ defined by

$f(n) = ne^{-\frac{\pi i n}{2}}.$

Determine whether $\{ f(n) \}$ converges or diverges, and if it converges find its limit.

This sequence diverges.
Proof. We saw in this exercise (Section 10.4, Exercise #20) that the sequence defined by $f(n) = e^{-\frac{\pi i n}{2}}$ diverges. We could use this to show that this sequence diverges (since for $n = 4k$ we have $f(n) = n$ and so $\{ f(n) \}$ cannot approach a finite limit). For practice, we can also prove this directly from the definition by contradiction as follows. Suppose there exists a real number $L$ and a positive integer $N$ such that for all $n > N$ and all $\varepsilon > 0$ we have

$|f(n) - L| < \varepsilon.$

Since $N$ is positive we know $4N > N$ and $4N+2 > N$ . So, letting $\varepsilon = \frac{1}{2}$ , we have

$\begin{align*} && | f(4N) - L | &< \frac{1}{2} & \text{and} && |f(4N+2) - L| &< \frac{1}{2} \\[9pt] \implies && | 4N - L | &< \frac{1}{2} & \text{and} && |-(4N+2) - L| &< \frac{1}{2} \\[9pt] && && \implies && |4N+2+L| &< \frac{1}{2}. \end{align*}$

Adding these two expressions and using the triangle inequality we have,

$\begin{align*} && |4N-L| + |4N+2+L| &< 1 \\[9pt] \implies && |4N - L + 4N + 2 + L| &< 1 \\[9pt] \implies && |8N + 2| &< 1 \\[9pt] \implies && 2 &< 1. \end{align*}$

This is a contradiction. Hence, the sequence $\{ f(n) \}$ does not tend to a limit $L$ .

Determine the convergence or divergence of f(n) = (1 / n) e^{-πin / 2}

by RoRi

February 28, 2016

Consider the function $f(n)$ defined by

$f(n) = \frac{1}{n} e^{-\frac{\pi i n}{2}}.$

Determine the convergence or divergence of the sequence $\{ f(n) \}$ and if it converges determine its limit.

This sequence converges since

$\begin{align*} \lim_{n \to \infty} |f(n)| &= \lim_{n \to \infty} \left| \frac{1}{n} \cdot e^{-\frac{\pi i n}{2}} \right| \\[9pt] &= \lim_{n \to \infty} \left( \frac{1}{n} \cdot \left| e^{-\frac{\pi i n}{2}} \right| \right). \end{align*}$

But each of the limits in the product exists since

$\lim_{n \to \infty} \frac{1}{n} = 0$

and

$\lim_{n \to \infty} \left| e^{-\frac{\pi i n}{2}} \right| = \lim_{n \to \infty} 1 = 1.$

Hence,

$\lim_{n \to \infty} |f(n)| = 0 \cdot 1 = 0.$

Therefore, $\lim_{n \to \infty} f(n) = 0$ (as we saw in this exercise, if $f(n)$ is complex valued and $\lim_{n \to \infty} |f(n)| = 0$ then the sequence $\lim_{n \to \infty} f(n) = 0$ as well). Hence, the sequence $\{ f(n) \}$ converges to 0.

Determine the convergence or divergence of f(n) = e^{-πin / 2}

by RoRi

February 28, 2016

Consider the function $f(n)$ defined by

$f(n) = e^{-\frac{\pi i n}{2}}.$

Determine whether the sequence $\{ f(n) \}$ converges or diverges, and if it converges determine its limit.

The sequence $\{ f(n) \}$ diverges.
Proof. First, we use DeMoivre’s theorem to write,

$\begin{align*} f(n) &= e^{-\frac{\pi i n}{2}} \\[9pt] &= \left( e^{-\frac{\pi i}{2}} \right)^n \\[9pt] &= \cos \left( \frac{-n \pi}{2} \right) - i \sin \left( \frac{n \pi}{2} \right). \end{align*}$

(Note: On page 380 Apostol claims (without proof) that the sequence defined by $f(n) = \sin \left( \frac{n \pi}{2} \right)$ is divergent. If you want to accept that then this sequence will diverge since a complex-valued sequence $f(n) = u(n) + i v(n)$ diverges if and only if the sequences defined by the functions $u(n)$ and $v(n)$ converge. Since it is a good exercise (and maybe Apostol wanted us to prove it ourselves) we can prove this is divergent from the definition.)

Suppose, for the sake of contradiction, that the sequence $\{ f (n) \}$ converges to a limit $L$ . Then we know for every $\varepsilon > 0$ there exists a positive integer $N$ such that for all $n > N$ we have

$|f(n) - L| < \varepsilon.$

Since $N$ is positive we know $4N > N$ and $4N+2 > N$ . Hence, taking $\varepsilon = \frac{1}{2}$ we have,

$\begin{align*} | f(4N) - L | &< \frac{1}{2} & \text{and} && |f(4N+2) - L| &< \frac{1}{2} \\[9pt] \implies |\cos (2N \pi) - i \sin (2N \pi) - L| &< \frac{1}{2} & \text{and} && |\cos (2N \pi + \pi) - i \sin (2N \pi + \pi) -L| &< \frac{1}{2} \\[9pt] \implies |1 - L| &< \frac{1}{2} & \text{and} && |1+L| &< \frac{1}{2}. \end{align*}$

Adding these two expressions and using the triangle inequality,

$\begin{align*} && |1-L| + |1+L| &< 1 \\[9pt] \implies && |1 - L + 1 + L| &< 1 \\ \implies && 2 < 1, \end{align*}$

a contradiction. Hence, the sequence $\{ f(n) \}$ diverges $. \qquad \blacksquare$

Determine the convergence or divergence of f(n) = (1 + i/2)^-n

by RoRi

February 28, 2016

Consider the function $f(n)$ defined by

$f(n) = \left( 1 + \frac{i}{2} \right)^{-n}.$

Determine whether the sequence $\{ f(n) \}$ converges or diverges, and if it converges find the limit.

First, we note that

$\left| 1 + \frac{1}{2}i \right| = \sqrt{1^2 + \left( \frac{1}{2} \right)^2} = \frac{\sqrt{5}}{2}.$

Then,

$\begin{align*} \left| \left( 1 + \frac{i}{2} \right)^{-n} \right| &= \left| \frac{1}{\left(1 + \frac{i}{2} \right)^n} \right| \\[9pt] &= \left| \left( \frac{1}{1+\frac{i}{2}} \right)^n \right|\\[9pt] &= \left| \frac{1}{1+\frac{i}{2}} \right|^n \\[9pt] &= \left(\frac{2}{\sqrt{5}}\right)^n. \end{align*}$

Therefore (since $\frac{2}{\sqrt{5}} < 1$ and is real) we know from (10.10) on page 380 of Apostol that

$\lim_{n \to \infty} |f(n)| = \lim_{n \to \infty} \left( \frac{2}{\sqrt{5}} \right)^n = 0.$

Hence, there exists an $N$ such that for all $n > N$ we have

$\lim_{n \to \infty} |f(n) - 0| < \varepsilon \qquad \text{for all } \varepsilon> 0.$

So $\lim_{n \to \infty} f(n) = 0$ , so the sequence $\{ f_n \}$ converges to 0.