Exercise 1

Use mathematical induction to prove that

1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}

is true for all positive integers $n$ .

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Solution

Let

P(n): 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}

where $n$ is a positive integer.

(1) Base Case:

Since

\frac{(1)(1+1)(2(1)+1)}{6} = \frac{(1)(2)(3)}{6} = 1 = 1^2,

$P(1)$ is true.

(2) Inductive Step:

Let $k$ be a positive integer.
Assume $P(k)$ is true:

1^2 + 2^2 + 3^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}.

We show that $P(k+1)$ is true:

1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 = \frac{(k+1)(k+2)(2k+3)}{6}.

Adding $(k+1)^2$ and substituting the inductive hypothesis:

= \frac{k(k+1)(2k+1)}{6} + (k+1)^2

= \frac{2k^3 + 3k^2 + k + 6(k^2 + 2k + 1)}{6}

= \frac{2k^3 + 3k^2 + k + 6k^2 + 12k + 6}{6}

= \frac{2k^3 + 9k^2 + 13k + 6}{6}

= \frac{(k+1)(k+2)(2k+3)}{6}.

Therefore, $P(k+1)$ is true.

By Mathematical Induction, $P(n)$ is true for all positive integers $n$ .

Exercise 2

Determine if the following sequences converge or diverge. If it converges, find its limit.

2.1 $\left\{ \left( \frac{2n+3}{2n-5} \right)^n \right\}$

2.2 $\{ \ln(n) - \ln(n+1) \}$

2.3 $\left\{ \frac{n^2}{2n-1} \sin\left(\frac{1}{n}\right) \right\}$

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Solution

2.1

Let

K = \left( \frac{2x+3}{2x-5} \right)^x.

Then

\ln K = x \ln \left( \frac{2x+3}{2x-5} \right).

\lim_{x \to \infty} \ln K = \lim_{x \to \infty} \frac{\ln\left( \frac{2x+3}{2x-5} \right)}{1/x}.

Using L’Hôpital’s Rule:

= \lim_{x \to \infty} \frac{\frac{2x-5}{2x+3} \left( \frac{(2x-5)(2)-(2x+3)(2)}{(2x-5)^2} \right)} {-1/x^2}

= \lim_{x \to \infty} \frac{2x-5}{2x+3} \left( \frac{4x-10-4x-6}{(2x-5)^2} \right)(-x^2)

= \lim_{x \to \infty} \frac{16x^2}{(2x+3)(2x-5)} = \lim_{x \to \infty} \frac{16x^2}{4x^2 - 4x - 15} = 4.

So,

\lim_{x \to \infty} K = e^4.

Therefore, the sequence converges to $e^4$ .

2.2

a_n = \ln(n) - \ln(n+1).

Let

f(x) = \ln \left( \frac{x}{x+1} \right).

\lim_{x \to \infty} f(x) = \ln \lim_{x \to \infty} \left( \frac{x}{x+1} \right) = \ln 1 = 0.

Therefore, the sequence converges to $0$ .

2.3

a_n = \frac{n^2}{2n-1} \sin\left( \frac{1}{n} \right).

Let

f(x) = \frac{x^2}{2x-1} \sin\left( \frac{1}{x} \right).

\lim_{x \to \infty} \frac{x}{2x-1} \left( \frac{\sin(1/x)}{1/x} \right).

\lim_{x \to \infty} \frac{\sin(1/x)}{1/x} = 1.

Thus,

\lim_{x \to \infty} \frac{x^2}{2x-1} \sin\left( \frac{1}{x} \right) = \frac{1}{2}.

Therefore, the sequence converges to $\frac{1}{2}$ .

Exercise 3

Consider the following sequences.
Are they monotone? bounded?
If monotone, determine whether increasing or decreasing.

3.1 $\left\{ \frac{1}{2^n} \right\}$

3.2 $\left\{ \frac{2^{n+1}}{n+2} \right\}$

3.3 $\{ 2n e^{-2n} \}$

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Solution

3.1

a_{n+1} - a_n = \frac{1}{2^{n+1}} - \frac{1}{2^n} = -\frac{1}{2^{n+1}} &lt; 0.

Thus the sequence is monotonic and decreasing.
It is bounded above and below.
Therefore, it is bounded.

3.2

\frac{a_{n+1}}{a_n} = \frac{2(n+2)}{n+3} &gt; 1.

Thus the sequence is monotonic and increasing.
Therefore, it is unbounded.

3.3

Let

f(x) = 2x e^{-2x}.

f'(x) = (-4x+1)e^{-2x} = \frac{-4x+1}{e^{2x}} &lt; 0.

Thus the sequence is monotonic and decreasing.
It is bounded above and below.
Therefore, it is bounded.

Exercise 4

Determine if the following infinite series converge or diverge.

4.1 Telescoping Series: $\sum_{n=1}^\infty \left( \frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)} \right)$

4.2 Geometric Series: $\sum_{n=1}^\infty \frac{3^{n-1}-1}{6^{n-1}}$

4.3 Geometric Series: $\sum_{n=0}^\infty \left( \frac{e}{\pi} \right)^n$

4.4 p-Series: $\sum_{n=1}^\infty \frac{1}{n^{5/4}}$

4.5 p-Series: $\sum_{n=1}^\infty \frac{n+1}{n^2 \sqrt{n}}$

4.6 $\sum_{n=1}^\infty \frac{1}{n^2+1}$ (Using Integral Test)

4.7 $\sum_{n=1}^\infty \frac{2}{1+e^n}$ (Using Integral Test)

4.8 $\sum_{n=1}^\infty \frac{\ln n}{n^3}$ (Using Comparison Test)

4.9 $\sum_{n=0}^\infty \frac{2^n+5}{3^n}$ (Using Ratio Test)

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Solution

4.1 (Telescoping Series)

S_n = -\frac{1}{\ln 2} + \frac{1}{\ln(n+2)}.

\lim_{x \to \infty} \left( -\frac{1}{\ln 2} + \frac{1}{\ln(x+2)} \right) = -\frac{1}{\ln 2}.

Convergent.

4.2

\sum \left( \frac{1}{2} \right)^{n-1} - \sum \left( \frac{1}{6} \right)^{n-1} = 2 - \frac{6}{5} = \frac{4}{5}.

4.3

Geometric series with

r = \frac{e}{\pi} &lt; 1.

\sum_{n=0}^\infty \left( \frac{e}{\pi} \right)^n = \frac{1}{1 - e/\pi} = \frac{\pi}{\pi - e}.

4.4

$p$ -series with $p = 5/4 > 1$ .
Convergent.

4.5

\sum \frac{1}{n^{3/2}} + \sum \frac{1}{n^{5/2}}.

Both converge.
Thus convergent.

4.6

\int_1^\infty \frac{1}{x^2+1} dx = \frac{\pi}{4}.

Convergent.

4.7

-2 \ln \left( \frac{e}{1+e} \right).

Convergent.

4.8

By comparison with $\frac{1}{n^2}$ , convergent.

4.9

L = \frac{2}{3} &lt; 1.

3 + \frac{15}{2} = \frac{21}{2}.

Convergent.

Exercise 5

Determine if the following alternating series converge or diverge?

5.1 $\sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{3n}{2n+1}$

5.2 $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n5^n}$

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Solution

5.1

Let

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

\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{3n}{2n+1} = \lim_{n \to \infty} \frac{3}{2 + 1/n} = \frac{3}{2}.

Since

\lim_{n \to \infty} a_n \neq 0,

the series

\sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{3n}{2n+1}

diverges.

5.2

Let

a_n = \frac{1}{n5^n}, \qquad a_{n+1} = \frac{1}{(n+1)5^{n+1}}.

Consider

\frac{1}{(n+1)5^{n+1}} &lt; \frac{1}{n5^n},

so the sequence is decreasing.

\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n5^n} = 0.

Therefore,

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n5^n}

converges by the Alternating Series Test.

Exercise 6

Determine if

\sum_{n=1}^{\infty} \frac{(-1)^n}{n!7^{-n}}

absolutely converge or conditionally converge or diverge. (Using Ratio Test)

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Solution

\sum_{n=1}^{\infty} \frac{(-1)^n}{n!7^{-n}} = \sum_{n=1}^{\infty} \frac{(-1)^n \cdot 7^n}{n!} = \sum_{n=1}^{\infty} \frac{(-7)^n}{n!}.

Let

a_n = \frac{(-7)^n}{n!}.

\left| \frac{a_{n+1}}{a_n} \right| = \frac{7^{n+1}}{(n+1)!} \cdot \frac{n!}{7^n} = \frac{7}{n+1}.

\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{7}{n+1} = 0.

Since $L = 0 < 1$ , the series absolutely converges.

Exercise 7

Determine if

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(3n+5)^2}

absolutely converge or conditionally converge or diverge. (Using Comparison Test)

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Solution

Let

a_n = \frac{(-1)^{n+1}}{(3n+5)^2}, \qquad b_n = \frac{1}{n^2}.

\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} \frac{1}{(3n+5)^2}.

Since

\frac{1}{(3n+5)^2} \le \frac{1}{n^2},

we have $|a_n| \le b_n$ .

\sum_{n=1}^{\infty} \frac{1}{n^2}

is a convergent $p$ -series with $p = 2 > 1$ .

Therefore,

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(3n+5)^2}

absolutely converges.

Exercise 8

Find the radius of convergence and the interval of convergence of the power series

\sum_{n=0}^{\infty} \frac{x^n}{3^{2n}}.

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Solution

\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{n+1}}{3^{2(n+1)}} \cdot \frac{3^{2n}}{x^n} \right| = \left| \frac{x}{3^2} \right| = \frac{|x|}{9}.

The series converges when

\frac{|x|}{9} &lt; 1 \quad \Rightarrow \quad |x| &lt; 9.

So

-9 &lt; x &lt; 9.

Endpoints:

At $x = 9$ :

\sum_{n=0}^{\infty} \frac{9^n}{3^{2n}} = \sum_{n=0}^{\infty} \frac{9^n}{9^n} = \sum_{n=0}^{\infty} 1,

which diverges.

At $x = -9$ :

\sum_{n=0}^{\infty} \frac{(-9)^n}{3^{2n}} = \sum_{n=0}^{\infty} (-1)^n,

which diverges.

Therefore, the radius of convergence is

R = 9

and the interval of convergence is

-9 &lt; x &lt; 9.

Exercise 9

Find the Taylor polynomial of $f(x) = \frac{1}{x}$ about the given point $x = 1$ .

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Solution

Let $f(x) = \frac{1}{x}$ , so $f(1) = 1$ .

f'(x) = -\frac{1}{x^2}, \quad f'(1) = -1

f''(x) = \frac{2}{x^3}, \quad f''(1) = 2

f^{(3)}(x) = -\frac{6}{x^4}, \quad f^{(3)}(1) = -6

f^{(4)}(x) = \frac{24}{x^5}, \quad f^{(4)}(1) = 24

f^{(5)}(x) = -\frac{120}{x^6}, \quad f^{(5)}(1) = -120

\cdots

f^{(n)}(x) = \frac{(-1)^n n!}{x^{n+1}}, \quad f^{(n)}(1) = (-1)^n n!

Therefore, the Taylor polynomial about $x = 1$ is:

f(1) + f'(1)(x-1) + \frac{1}{2!} f''(1)(x-1)^2 + \frac{1}{3!} f^{(3)}(1)(x-1)^3 + \dots + \frac{1}{n!} f^{(n)}(1)(x-1)^n

= 1 + (-1)(x-1) + \frac{1}{2!}(2)(x-1)^2 + \frac{1}{3!}(-6)(x-1)^3 + \dots + \frac{1}{n!} (-1)^n(n!)(x-1)^n

= 1 - (x-1) + (x-1)^2 - (x-1)^3 + (x-1)^4 - (x-1)^5 + \dots + (-1)^n (x-1)^n.

Exercise 10

Find Maclaurin series of $f(x) = \ln(1-x)$ .

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Solution

Let $f(x) = \ln(1-x)$ , so $f(0) = 0$ .

f'(x) = -\frac{1}{1-x} = (-1)(1-x)^{-1}, \quad f'(0) = -1

f''(x) = (-1)(1-x)^{-2}, \quad f''(0) = -1

f^{(3)}(x) = (-1)(-2)(1-x)^{-3}(-1), \quad f^{(3)}(0) = -2

f^{(4)}(x) = (-1)(-2)(-3)(1-x)^{-4}(-1)(-1), \quad f^{(4)}(0) = -3!

\cdots

f^{(n)}(x) = (-1)^{2n-1}(n-1)!(1-x)^{-n} = -(n-1)!(1-x)^{-n}, \quad f^{(n)}(0) = -(n-1)!

Therefore,

\sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=1}^{\infty} \frac{-(n-1)!}{n!} x^n = -\sum_{n=1}^{\infty} \frac{x^n}{n}.

Exercise 11

Find the power series of $f(x) = \frac{1}{\sqrt{4-x}}$ .

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Solution

f(x) = \frac{1}{\sqrt{4-x}} = \frac{1}{2(1 - x/4)^{1/2}} = \frac{1}{2}(1 - \frac{x}{4})^{-1/2}.

Using

(1+x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n,

take $k = -\frac{1}{2}$ and substitute $x \mapsto -\frac{x}{4}$ .

\binom{-1/2}{n} = \frac{(-1/2)(-3/2)(-5/2)\dots(-1/2-n+1)}{n!} = \frac{(-1)^n (1\cdot3\cdot5\cdots(2n-1))}{2^n n!}.

Thus,

\frac{1}{2}(1 - \frac{x}{4})^{-1/2} = \frac{1}{2} \sum_{n=0}^{\infty} \binom{-1/2}{n} \left(-\frac{x}{4}\right)^n

= \frac{1}{2} \sum_{n=0}^{\infty} \left[ \frac{(-1)^n (1\cdot3\cdot5\cdots(2n-1))}{2^n n!} \cdot \frac{(-1)^n x^n}{4^n} \right]

= \frac{1}{2} \sum_{n=0}^{\infty} \frac{(1\cdot3\cdot5\cdots(2n-1)) x^n}{8^n n!}

= \frac{1}{2} \left[ 1 + \frac{x}{8} + \frac{3x^2}{8^2 \cdot 2!} + \frac{15x^3}{8^3 \cdot 3!} + \dots \right].

Exercise 12

Draw graphs and write the Fourier series of

f(x) = \begin{cases} x+2, & -2 \le x &lt; 0 \\ 2, & 0 &lt; x \le 2 \end{cases}

Show solution ↓

Solution

Since $2L = 4$ , we have $L = 2$ .

f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right).

Finding $a_0$ :

a_0 = \frac{1}{L} \int_{-L}^{L} f(x)\,dx = \frac{1}{2} \int_{-2}^{2} f(x)\,dx

= \frac{1}{2} \left( \int_{0}^{2} 2\,dx + \int_{-2}^{0} (x+2)\,dx \right)

= \frac{1}{2} \left( [2x]_{0}^{2} + \left[\frac{x^2}{2}+2x\right]_{-2}^{0} \right) = 3.

Finding $a_n$ :

a_n = \frac{1}{2} \left( \int_{0}^{2} 2\cos\left(\frac{n\pi x}{2}\right)dx + \int_{-2}^{0} (x+2)\cos\left(\frac{n\pi x}{2}\right)dx \right)

= \frac{2}{(n\pi)^2}(1 - (-1)^n) = -\frac{2}{(n\pi)^2}((-1)^n - 1).

Finding $b_n$ :

b_n = \frac{1}{2} \left( \int_{0}^{2} 2\sin\left(\frac{n\pi x}{2}\right)dx + \int_{-2}^{0} (x+2)\sin\left(\frac{n\pi x}{2}\right)dx \right)

= -\frac{2}{n\pi} (-1)^n.

Therefore, the Fourier series is:

f(x) = \frac{3}{2} + \sum_{n=1}^{\infty} \left( -\frac{2}{(n\pi)^2}((-1)^n - 1) \cos\left(\frac{n\pi x}{2}\right) - \frac{2}{n\pi} (-1)^n \sin\left(\frac{n\pi x}{2}\right) \right).

Exercise 13

Write Fourier series of $f(x) = |x| - 1$ with period $-2 \le x \le 2$ .

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Solution

From

f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right).

Since the period is $2L = 4$ , we have $L = 2$ .

Check parity:

f(-x) = |-x| - 1 = |x| - 1 = f(x).

Thus, $f(x)$ is even, so $b_n = 0$ .

Find $a_0$ :

a_0 = \frac{1}{L} \int_{-L}^{L} f(x)\,dx = \frac{1}{2} \int_{-2}^{2} (|x| - 1)\,dx = \int_0^2 (x - 1)\,dx.

= \left[ \frac{x^2}{2} - x \right]_0^2 = (2 - 2) - 0 = 0.

Find $a_n$ :

a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{2}\right) dx = \int_0^2 (x - 1) \cos\left(\frac{n\pi x}{2}\right) dx.

Integrating by parts,

a_n = \frac{4}{(n\pi)^2} \left( (-1)^n - 1 \right).

Thus, the Fourier series is

f(x) = \sum_{n=1}^\infty \frac{4}{(n\pi)^2} \left( (-1)^n - 1 \right) \cos\left(\frac{n\pi x}{2}\right).

Exercise 14

Write Fourier series of

f(x) = \begin{cases} 2, & -2 &lt; x &lt; 0 \\ -2, & 0 &lt; x \le 2 \end{cases}

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Solution

Here $2L = 4$ , so $L = 2$ .

Check parity:

f(-x) = -f(x),

so $f(x)$ is odd. Hence $a_0 = a_n = 0$ .

Find $b_n$ :

b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{2}\right) dx = \frac{2}{2} \int_0^2 (-2) \sin\left(\frac{n\pi x}{2}\right) dx.

= -2 \int_0^2 \sin\left(\frac{n\pi x}{2}\right) dx = \frac{4}{n\pi} \left( (-1)^n - 1 \right).

Thus,

f(x) = \sum_{n=1}^\infty \frac{4}{n\pi} \left( (-1)^n - 1 \right) \sin\left(\frac{n\pi x}{2}\right).

Exercise 15

Determine Fourier series of

f(x) = \begin{cases} x+\pi, & -\pi \le x &lt; 0 \\ x-\pi, & 0 &lt; x \le \pi \end{cases}

and show that

\sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n-1} = \frac{\pi}{4}.

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Solution

Here $2L = 2\pi$ , so $L = \pi$ .

Check parity:

f(-x) = -f(x),

so $f(x)$ is odd. Hence $a_0 = a_n = 0$ .

Find $b_n$ :

b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx)\,dx = \frac{2}{\pi} \int_0^\pi (x-\pi) \sin(nx)\,dx.

After integration by parts,

b_n = -\frac{2}{n}.

Thus,

f(x) = -\sum_{n=1}^\infty \frac{2}{n} \sin(nx).

Evaluate at $x = \frac{\pi}{2}$ :

-\frac{\pi}{2} = -2 \sum_{n=1}^\infty \frac{1}{n} \sin\left(\frac{n\pi}{2}\right).

Hence,

\frac{\pi}{4} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n-1}.

Exercise 16

Determine Fourier series of

f(x) = \begin{cases} x+2, & -2 &lt; x \le 0 \\ 0, & 0 &lt; x \le 2 \end{cases}

and find the convergent value at $x = 0$ .

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Solution

Here $2L = 4$ , so $L = 2$ .

Find $a_0$ :

a_0 = \frac{1}{2} \int_{-2}^2 f(x)\,dx = \frac{1}{2} \int_{-2}^0 (x+2)\,dx.

= \frac{1}{2} \left[ \frac{x^2}{2} + 2x \right]_{-2}^0 = 1.

Find $a_n$ :

a_n = \frac{1}{2} \int_{-2}^0 (x+2) \cos\left(\frac{n\pi x}{2}\right) dx = -\frac{2}{(n\pi)^2} \left( (-1)^n - 1 \right).

Find $b_n$ :

b_n = \frac{1}{2} \int_{-2}^0 (x+2) \sin\left(\frac{n\pi x}{2}\right) dx = -\frac{2}{n\pi}.

Thus,

f(x) = \frac{1}{2} + \sum_{n=1}^\infty \left( -\frac{2}{(n\pi)^2} \left( (-1)^n - 1 \right) \cos\left(\frac{n\pi x}{2}\right) - \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right).

Since $x=0$ is a point of discontinuity, the Fourier series converges to

\frac{f(0^+) + f(0^-)}{2} = \frac{0 + 2}{2} = 1.

Exercise Set 2

Theory

Exercises