1 Solution 1
\((\dfrac{50n}{n^2 - 8n + 17})_{n=1}^{\infty}\)
Let \(f(n) = \dfrac{50n}{n^2 - 8n + 17}\)
\(f(1) = \dfrac{50}{1^2 - 8 + 17} = \dfrac{50}{10} = 5\)
\(f(2) = \dfrac{50.2}{2^2 - 16 + 17} = \dfrac{50.2}{4+1} = \dfrac{50.2}{5} = 20\)
\(f(n) = \dfrac{50n}{(n-4)^2 + 1}\)
\(f(3) = \dfrac{50.3}{1^2 + 1} = \dfrac{50.3}{2} = 25.3 = 75\)
\(f(4) = \dfrac{50.4}{1} = 200\)
From theorem 2.8.2, \(\lim_{n \to \infty} \dfrac{50n}{(n-4)^2 + 1}\)
\(= \lim_{n \to \infty} \dfrac{50n}{n^2(1 - \dfrac{4}{n})^2 + 1}\)
\(= \lim_{n \to \infty} \dfrac{50}{n((1-\dfrac{4}{n})^2 + \dfrac{1}{n})}\)
\(= 0\)
2 Solution 2
\((\dfrac{1 + (-1)^n}{n})_{n=1}^{\infty}\)
Let \(f(n) = \dfrac{1 + (-1)^n}{n}\)
\(f(1) = \dfrac{1 + (-1)^1}{1} = \dfrac{1-1}{1} = 0\)
\(f(2) = \dfrac{1 + (-1)^2}{2} = \dfrac{1+1}{2} = 1\)
\(f(3) = \dfrac{1 + (-1)^3}{3} = \dfrac{1-1}{3} = 0\)
\(f(4) = \dfrac{1 + (-1)^4}{4} = \dfrac{1+1}{4} = \frac{1}{2}\)
From theorem 2.8.2, \(\lim_{n \to \infty} \dfrac{1 + (-1)^n}{n}\)
\(= \lim_{n \to \infty} \dfrac{1}{n} + \lim_{n \to \infty} \dfrac{-1^n}{n}\)
\(= 0 + 0 = 0\)
3 Solution 3
\((sin (\pi n))_{n=3}^{\infty}\)
Let \(f(n) = sin (\pi n)\)
\(f(3) = sin 3\pi = sin \pi = 0\)
\(f(4) = sin 4\pi = sin 0 = 0\)
\(f(5) = sin 5\pi = sin \pi = 0\)
\(f(6) = sin 6\pi = sin 0 = 0\)
\(\lim_{n \to \infty} sin (\pi n)\)
We know that \(n\) starts from \(3\).
Since function is periodic at \(2\pi\),
\(\lim_{n \to \infty} sin (\pi n) = 0\)
4 Solution 4
\(\lim_{n \to \infty} \dfrac{3n^2 - 1}{n^2 + 3}\)
From theorem 2.8.2,
\(\lim_{n \to \infty} \dfrac{3n^2 - 1}{n^2 + 3} = \dfrac{3-0}{1+0} = 3\)
5 Solution 5
\(\lim_{n \to \infty} (\dfrac{n^2}{n+1} + \dfrac{4-n^3}{n^2-1})\)
From theorem 2.8.2,
\(\lim_{n \to \infty} (\dfrac{n^2}{n+1} + \dfrac{4-n^3}{n^2-1})\)
\(= \lim_{n \to \infty} (\dfrac{n^2(n-1) + (4-n^3)}{(n+1)(n-1)}\)
\(= \lim_{n \to \infty} \dfrac{4-n^2}{(n+1)(n-1)}\)
\(= \lim_{n \to \infty} \dfrac{\dfrac{4}{n^2} - 1}{(1 + \dfrac{1}{n})(1-\dfrac{1}{n})}\)
\(= \dfrac{-1}{(1+0)(1-0)} = -1\)
6 Solution 6
\(\lim_{n \to \infty} (n^3(\dfrac{1}{n^2} - \dfrac{1}{(n+1)^2}))\)
From theorem 2.8.2,
\(\lim_{n \to \infty} (n^3(\dfrac{(n+1)^2}{n^2(n+1)^2} - \dfrac{n^2}{n^2(n+1)^2}))\)
\(= \lim_{n \to \infty} (n^3(\dfrac{(n+1)^2 - n^2}{n^2(n+1)^2}\)
\(= \lim_{n \to \infty} (n(\dfrac{n^2 + 1 + 2n - n^2}{(n+1)^2}\)
\(= \lim_{n \to \infty} (n(\dfrac{(1+2n)}{(n+1)^2}\)
\(= \lim_{n \to \infty} ((\dfrac{n^2(1/n+2)}{n^2(1+1/n)^2}\)
\(= \dfrac{0+2}{1+0} = 2\)
7 Solution 7
\(\lim_{n \to \infty} \dfrac{n \sin n}{n^2 + 1}\)
From theorem 2.8.2,
\(\lim_{n \to \infty} \dfrac{n}{n^2(1 + \dfrac{\sin n}{n^2})}\)
\(= \lim_{n \to \infty} \dfrac{1}{n(1 + \dfrac{\sin n}{n^2})}\)
\(= 0\)
8 Solution 8
\(\lim_{n \to \infty} \dfrac{n}{n^2 + \sin n}\)
From theorem 2.8.2,
\(\lim_{n \to \infty} \dfrac{n}{n^2(1 + \sin n/n^2)}\)
\(= 0\)
9 Solution 9
\(\lim_{n \to \infty} \dfrac{1}{(3 + (-1)^n)^n}\)
When \(n\) is even, \((-1)^n\) is positive one.
When \(n\) is odd, \((-1)^n\) is negative one.
But in both cases,
\((3 + (-1)^n)\) is positive.
When \(n\) is even, the value is \(\dfrac{1}{4^n} = \dfrac{1}{2^{2n}}\)
When \(n\) is odd, the value is \(\dfrac{1}{2^n}\)
We can observe that as \(n \to \infty\), the function tends to zero.
\(\lim_{n \to \infty} \dfrac{1}{2^n} = 0\)
\(\lim_{n \to \infty} \dfrac{1}{2^{2n}} = 0\)
10 Solution 10
\(\lim_{n \to \infty} \dfrac{1}{(2 + (-1)^n)^n} = 0\)
When \(n\) is even, \((-1)^n\) is positive.
When \(n\) is odd, \((-1)^n\) is negative.
Overall, \(2 + (-1)^n\) can be either \(3\) or \(1\) based on either \(n\) is even or odd.
When \(n\) is even,
\(\lim_{n \to \infty} \dfrac{1}{3^n}\)
Let \(u = f(n) = 3^n\)
As \(n \to \infty\), \(u \to \infty\)
\(\lim_{u \to \infty} \dfrac{1}{u} = 0\)
When \(n\) is odd,
\(\lim_{n \to \infty} \dfrac{1}{1^n} = 1\)
So, the limit is undefined as it oscillates between \(0\) and \(1\).
11 Solution 11
\(\lim_{n \to \infty} \dfrac{1}{2^n + (-1)^n}\)
When \(n\) is even, \((-1)^n\) is 1.
When \(n\) is odd, \((-1)^n\) is \(-1\).
When \(n\) is even,
\(\lim_{n \to \infty} \dfrac{1}{2^n + 1}\)
Let \(u = f(n) = 2^n + 1\)
As \(n \to \infty\), \(u \to \infty\)
\(\lim_{u \to \infty} \dfrac{1}{u} = 0\)
When \(n\) is odd,
\(\lim_{n \to \infty} \dfrac{1}{2^n - 1}\)
Let \(u = f(n) = 2^n - 1\)
As \(n \to \infty\), \(u \to \infty\)
As \(u \to \infty, \dfrac{1}{u} \to 0\)
So, \(\lim_{n \to \infty} \dfrac{1}{2^n - 1} = 0\)
So, \(\lim_{n \to \infty} \dfrac{1}{2^n + (-1)^n} = 0\)
12 Solution 12
\(\lim_{n \to \infty} (\sqrt{n^3 + n} - \sqrt[n]{n})\)
\(= \lim_{n \to \infty} n(\sqrt{n + \dfrac{1}{n}} - \sqrt[n]{\dfrac{1}{n-1}})\)
\(= \infty\)
13 Solution 13
\(\lim_{n \to \infty} (\sqrt{n^3 + n^2} - \sqrt[n]{n})\)
\(= \lim_{n \to \infty} n(\sqrt{n + 1} - \sqrt[n]{\dfrac{1}{n-1}})\)
\(= \infty\)
14 Solution 14
\(\lim_{n \to \infty} \sin (\dfrac{(2n-1)\pi}{2})\)
\(= \lim_{n \to \infty} \sin (n\pi - \dfrac{\pi}{2})\)
\(= -\lim_{n \to \infty} \cos n\pi\)
This means that the limit value will be oscillating between \(1\) and \(-1\). So it is undefined.
15 Solution 15
\(\lim_{n \to \infty} \sin (\dfrac{(4n-1)\pi}{2})\)
\(= \lim_{n \to \infty} \sin (2n\pi - \dfrac{\pi}{2})\)
\(= - \lim_{n \to \infty} \cos 2n\pi\)
Since \(\cos\) function is periodic at \(2\pi\),
\(- \lim_{n \to \infty} \cos 2n\pi = -1\)
16 Solution 16
\(\lim_{n \to \infty} \sin (\dfrac{6\pi n^2}{3n+1})\)
Let \(u = f(n) = \dfrac{6\pi n^2}{3n+1}\)
\(\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \dfrac{n^2(6\pi)}{n^2(\dfrac{3}{n} + \dfrac{1}{n^2})}\)
\(= \lim_{n \to \infty} \dfrac{6\pi}{\dfrac{3}{n} + \dfrac{1}{n^2}} = \infty\)
As \(n \to \infty\), \(u \to \infty\)
\(\lim_{u \to \infty} \sin u\)
The limit diverges and hence it is not defined.
17 Solution 17
\(\forall n \in \mathbb{Z} 3^n > n^2 + 1\)
We will use mathematical induction.
Base case: If \(n=1\), \(3^1 = 3 > 1^2 + 1 = 2\)
Induction step: Suppose \(n\) is a postive integer and suppose \(3^n > n^2 + 1\).
\(3^{n + 1} = 3.3^n > 3(n^2 + 1)\)
\(3(n^2 + 1) = 3n^2 + 3 = n^2 + 2 + 1 + n^2\)
Since \(\forall n \in Z\), \(n^2 + 1 \geq 2n\)
So, \(n^2 + 2 + 1 + n^2 \geq n^2 + 2 + 2n\)
\(n^2 + 2 + 1 + n^2 \geq (n+1)^2 + 1\)
So, \(3^n+1 \geq (n+1)^2 + 1\)
18 Solution 18
18.1 Solution a
\(\forall n \in \mathbb{Z} (\dfrac{3}{2})^n > n\)
Let's prove for \(\forall n \geq 2 \in \mathbb{Z} (\dfrac{3}{2})^n > n\)
We will use mathematical induction.
Base case: \(n=2\)
\((\dfrac{3}{2})^n = \dfrac{9}{4} = 2.25 > 2\)
Induction step: Suppose \(n\) is a postive integer \(\geq 2\) and suppose \((\dfrac{3}{2})^n > n\).
\((\dfrac{3}{2})^{n+1} = \dfrac{3}{2}.(\dfrac{3}{2})^n > (\dfrac{3}{2})^n\)
\(\dfrac{3n}{2} = n + \dfrac{n}{2}\)
We know that \(\forall n \geq 2, \dfrac{n}{2} \geq 1\)
So, \(n + \dfrac{n}{2} \geq n + 1\)
\(\dfrac{3n}{2} \geq n + 1\)
So, \((\dfrac{3}{2})^{n+1} > n + 1\)
Now let's handle the case for \(n=1\),
\((\dfrac{3}{2})^1 = 1.5 > 1\)
18.2 Solution b
\(\lim_{n \to \infty} \dfrac{2^n}{3^n}\)
From (a), we know that \((\dfrac{3}{2})^2 > n\)
\((\dfrac{2}{3})^n < \dfrac{1}{n}\)
Since \(0 < \dfrac{2}{3}\),
\(\forall n \geq 1, 0 < (\dfrac{2}{3})^n < \dfrac{1}{n}\)
We know that \(\lim_{n \to \infty} \dfrac{1}{n} = 0\)
From squeeze theorem, \(\lim_{n \to \infty} \dfrac{2^n}{3^n} = 0\)
19 Solution 19
\((a_n)_{n=1}^{\infty}\)
\(a_1 = \dfrac{1}{3}, a_{n+1} = \dfrac{1-a_n}{3-4a_n}\)
We need to prove \(a_n = \dfrac{n}{2n+1}\)
We will use mathematical induction.
Base case: \(n = 1\)
\(a_n = a_1 = \dfrac{1}{3} = \dfrac{1}{2.1 + 1} = \dfrac{n}{2n + 1}\)
Induction step: Suppose \(n\) is a postive integer and \(a_n = \dfrac{n}{2n + 1}\).
\(a_{n+1} = \dfrac{1-a_n}{3-4a_n}\)
\(= \dfrac{1 - \dfrac{n}{2n+1}}{3-\dfrac{4n}{2n+1}}\)
\(= \dfrac{2n+1-n}{6n+3-4n}\)
\(= \dfrac{1+n}{2n+3}\)
\(= \dfrac{n+1}{2(n+1) + 1}\)
20 Solution 20
\((a_n)_{n=1}^{\infty}\)
\(a_1 = 1, a_{n+1} = (\dfrac{a_n}{n} + 1)^2\)
Now we need to find a formula for \(a_n\)
Let's see how the sequence goes.
\(a_1 = 1, a_2 = 4, a_3 = 9, a_4 = 16, a_5 = 25 ...\cdot\)
Looking at the series, I can guess that \(a_n = n^2\)
Let's try to prove it. We will use mathematical induction.
Base case: \(n = 1\)
\(a_1 = 1 = 1^2 = n^2\)
Induction step: Suppose \(n\) is a postive integer and \(a_n = n^2\).
\(a_{n+1} = (\dfrac{a_n}{n} + 1)^2\)
\(= (\dfrac{n^2}{n} + 1) = (n+1)^2\)
21 Solution 21
21.1 Solution a
\(\forall m > 0 \forall n > 0 (m+n)! > m^n\)
Suppose \(m > 0\)
Suppose \(n > 0\)
We will use mathematical induction on \(n\).
Base case: \(n=1\)
\((m+1)! = (m +1)m!\)
\((m+1)m! > m\)
So, \((m+1)! > m^n\)
Induction step: Suppose \(n\) is a postive integer and \((m+n)! = m^n\).
\((m+n+1)! = (m+n)!(m+n+1) > (m+n+1)m^n\)
Since \(m>0\) and \(n>0\),
\(m + n + 1 > m\)
So, \((m+n+1)m^n > m.m^n\)
\((m+n+1)m^n > m^{n+1}\)
So, \((m+n+1)! > m^{n+1}\)
21.2 Solution b
\(\forall k > 0 (2k^2)! > k^{2k^2}\)
Suppose \(a = k^2\)
\(k^{2k^2} = (k^2)^{k^2} = a^a\)
So we will try to prove \((2a)! > a^a\)
It can be re-written as \((a + a)! > a^a\)
From (a), with \(m = a\) and \(n=a\), we can conclude that \((a+a)! > a^a\)
Hence \(\forall k > 0 (2k^2)! > k^{2k^2}\)
21.3 Solution c
\(\forall k > 0\; \exists N > 0\; \forall n \geq N\; n! > k^n\)
Suppose \(k > 0\)
Let \(N = 2k^2\)
Suppose \(n \geq N\)
We will use mathematical induction.
Base case: \(n = 2k^2\)
\(n! = (2k^2)! > k^{2k^2}\) (From (b))
Induction step: Suppose \(n! > k^n\)
\((n+1)! = (n+1)n! > (n+1)k^n\)
We know that \(2k^2 + 1 > k\)
Since \(n \geq N\), \(n \geq 2k^2\)
\(n + 1 \geq 2k^2 + 1 > k\)
So, \(n + 1 > k\)
\((n+1)k^n > k^{n+1}\)
So, \((n+1)! > k^{n+1}\)
22 Solution 22
Suppose \(\lim_{x \to \infty} f(x) = L\)
So, \(\forall \epsilon > 0 \; \exists N \; x > N \implies |f(x)-L| < \epsilon\)
Now let's prove \(\lim_{n \to \infty} f(n) = L\)
Suppose \(\epsilon > 0\). There exists some \(N\) such that \(n > N\) as \(n\) belongs to the domain of the function \(f\). From our earlier assumption, we can conclude that \(|f(n) - L| < \epsilon\)
23 Solution 23
We need to find \(f\) and a nuber \(a\) such that for all \(x \neq a, f(x) > 0\), but \(\lim_{x \to a}f(x) = 0\).
\(a = 0\)
\(f(x) = x^2\)
To see why we strict inequalities aren't preserved by limits:
Let \(r=2\)
\(0 < |x-a| < d\)
Since \(|x-a| > 0\) and \(a = 0\), let's assume \(x = 2\)
\(f(x) = 4 > r\).
But \(L = \lim_{x \to a}f(x) = 0 > r\) is not true since \(r\) is \(2\).
So we can see that strict inequalities aren't preserved by limits.
24 Solution 24
Let \(\lim_{n \to \infty} a_n = L\)
We need to prove \(\exists M \; \forall n > 0 \; |a_n| \leq M\)
Suppose \(\epsilon < 1\)
Suppose \(n\) is a positive integer.
We know that \(|f(n) - L| < 1\)
Since \(f(n) = a_n\)
\(|a_n - L| < 1\)
We know that \(|x-y| \geq |x| - |y|\)
So, \(|a_n - L| \geq |a_n| - |L|\)
\(|a_n| - |L| \leq |a_n -L|\)
\(|a_n| - |L| \leq |a_n -L| < 1\)
\(|a_n| - |L| \leq 1\)
\(|a_n| \leq |L| + 1\)
So, there exists \(M\) such that \(|a_n| \leq M\)
25 Solution 25
25.1 Solution a
Let us prove by contradiction.
So, let's assume that in the nested interval theorem we can strengthen the conclusion.
So, \(\forall n \; u_n < c < v_n\)
Let's consider two sequences \((u_n)_{n=1}^\infty\) and \((v_n)_{n=1}^\infty\) such that \(u_n = 0\) and \(v_n = 1/n\).
\(\lim_{n \to \infty} u_n = c\) and \(\lim_{n \to \infty} v_n = c\)
But \(c < c < c\) isn't true. Hence our initial assumptions isn't right.
25.2 Solution b
Suppose \(u_1 < u_2 < u_3 < ...\cdot\)
Suppose \(v_1 < v_2 < v_3 < ...\cdot\)
Suppose \(\forall n \; u_n < v_n\)
Suppose \(\lim_{n \to \infty} u_n - v_n = 0\)
From nested interval theorem, we know that \(\forall n \; u_n \leq c \leq v_n\) and \(\lim_{n \to \infty} u_n = c\), \(\lim_{n \to \infty} v_n = c\).
We know that \(u_n < v_n\)
That means either \(u_n \leq c < v_n\) or \(u_n < c \leq v_n\)
Let's take the first case \(u_n \leq c < v_n\)
Let's try to prove \(u_n < c\) by contradiction.
If there is a \(k \in \mathbb{N}\) such that \(u_k = c\), then \(c = u_k < u_{k+1}\).
This contradicts the assumption that \(\forall n \; u_n \leq c \leq v_n\). Similar argument holds for \(u_n < c \leq v_n\).
Hence \(u_n < c < v_n\).