Proposition 2. Comparison Rules. Suppose \(\lim _{n \to \infty } a_n \) and \(\lim _{n \to \infty } b_n\) exist, then the following implications hold:
We first prove (a). Let \(\lim _{n \to \infty } a_n = L\) and \(\lim _{n \to \infty } b_n = M\). Given that sequences \(a_n, b_n\) have a finite limit and \(L<M\). The key idea is to bound \(a_n\) and \(b_n\) in two disjoint intervals: \((L-\epsilon , L + \epsilon )\) and \((M-\epsilon , M+\epsilon )\). Picking \(\epsilon = \frac {M-L}{2} >0\), we start the formal proof
Proof. \begin {align*} \because a_n \to L, \forall \epsilon >0, \exists N_1>0 \text { s.t } \forall n > N_1\\ |a_n-L| < \epsilon \\ \implies L- \epsilon <a_n<L+\epsilon \end {align*}
Similarly, \begin {align*} \because b_n \to L, \forall \epsilon >0, \exists N_2>0 \text { s.t } \forall n > N_2\\ |b_n-M| <\epsilon \\ \implies M -\epsilon <b_n<M+\epsilon \end {align*}
Note that \(M-\frac {M-L}{2}=L+\frac {M-L}{2} = \frac {M+L}{2}\). Combining both results, for \(n > \max \{N_1,N_2\}\), \begin {align*} a_n < \frac {M+L}{2} < b_n \end {align*}
Hence, (a) is proven. Now, (b) is simply just the contrapositive statement of (a), we have to show that (b) is indeed the contrapositive of (a). \begin {align*} \neg (\exists N>0 \text { s.t } a_n < b_n \forall n>N)\\ \iff \forall N>0, \exists n >N \text { s.t } a_n \geq b_n \end {align*}
The statement is equivalent to there being infinitely many \(n\) such that \(a_n\geq b_n\). Naturally, the statement \(\lim _{n \to \infty }a_n \geq \lim _{n \to \infty }b_n\) is the negation of the statement \(\lim _{n \to \infty } a_n < \lim _{n \to \infty } b_n \). Hence (b) is indeed a contrapositive. □
Theorem 6 (Ratio Test for sequences). Suppose \(\{a_n\}\) is a sequence of positive numbers such that \begin {align*} \lim _{n\to \infty } \frac {a_{n+1}}{a_n} = L \end {align*}
Where \(0\leq L < 1\), then \(\lim _{n \to \infty }a_n =0\)
Proof. Here, we need to use comparison proposition (a) to prove this theorem. Because \(0 \leq L<\frac {L+1}{2}<1\), \begin {align*} \lim _{n\to \infty }\frac {a_{n+1}}{a_n} = L < \frac {L+1}{2}\\ \implies \exists N > 0 \text { s.t } \frac {a_{n+1}}{a_n} < \frac {L+1}{2} \forall n>N \end {align*}
Note that for \(n>N\), \begin {align*} a_n = \frac {a_n}{a_{n-1}}\cdot \frac {a_{n-1}}{a_{n-2}}\cdot ... \frac {a_{N+2}}{a_{N+1}}\cdot a_{N+1} \end {align*}
Each fraction above is less than \(\frac {L+1}{2}\), and counting carefully, we have \begin {align*} a_n < \left (\frac {L+1}{2}\right )^{n-N-1}a_{N+1} = \left (\frac {L+1}{2}\right )^n\left (\frac {L+1}{2}\right )^{-(N+1)}a_{N+1} \end {align*}
Because, \(\left (\frac {L+1}{2}\right )^n \to 0\) and \(0\leq a_n\), by the squeeze theorem, \(a_n \to 0\) □
Theorem 7 (Root Test for sequences). Suppose \(\{a_n\}\) is a sequence of real numbers such that \begin {align*} \lim _{n \to \infty } \sqrt [n]{|a_n|} = L \end {align*}
Where \(0\leq L < 1\), then \(\lim \limits _{n \to \infty }a_n =0\).
Proof. Again, using the fact \begin {align*} 0\leq L < \frac {L+1}{2} < 1 \end {align*}
and comparison of limits proposition (a), \begin {align*} \because \lim _{n \to \infty } \sqrt [n]{|a_n|} = L < \frac {L+1}{2}\\ \exists N>0 \text { s.t. } \forall n >N \quad \sqrt [n]{|a_n|} < \frac {L+1}{2} \\ \implies |a_n| < \left ( \frac {L+1}{2}\right )^n \end {align*}
Naturally, \(0 \leq |a_n|\), and by the fact that \(\left ( \frac {L+1}{2}\right )^n \to 0\), the squeeze theorem applies and \(|a_n| \to 0\), and \(a_n \to 0\) well: \begin {align*} \because -|a_n| \leq a_n \leq |a_n| \end {align*}
If \(|a_n|\) converges to 0, \(-|a_n|\) also converges to 0, hence by the squeeze theorem, \(a_n \to 0\) □
Proposition 3. Any polynomial \(p(n)\) with a positive leading term, is positive for sufficiently large \(n\)
Proof. Let the polynomial be of degree \(k\). Then we have \begin {align*} p(n) = a_kn^k + a_{k-1}n^{k-1}+...+a_1n + a_0 \end {align*}
With \(a_k >0\). Dividing by \(n^k\), we have: \begin {align*} \frac {p(n)}{n^k} = a_k + \frac {a_{k-1}}{n} + ... + \frac {a_1}{n^{k-1}} + \frac {a_0}{n^k} \\ \implies \lim _{n \to \infty }\frac {p(n)}{n^k} = a_k > 0\\ \implies \exists N > 0 \text { s.t } \frac {p(n)}{n^k} > 0 \quad \forall n >N \end {align*}
From this we get the result that \(p(n)>0\) for all \(n\) greater than some \(N\). □
Proposition 4. Given two sequences \(\{x_n\}\) and \(\{y_n\}\) of positive terms, and there exists some \(N \in \mathbb {N}\) such that whenever \(n\geq N\), we have \begin {align*} \frac {x_{n+1}}{x_n} \leq \frac {y_{n+1}}{y_n} \end {align*}
Then, if \(\lim \limits _{n \to \infty } y_n =0\), we also have \(x_n \to 0\).
Proof. If we have \(\frac {x_{n+1}}{x_n} \leq \frac {y_{n+1}}{y_n}\) for an infinite number of terms, then we have: \begin {align*} \lim _{n \to \infty }\frac {x_{n+1}}{x_n} \leq \lim _{n \to \infty }\frac {y_{n+1}}{y_n} \end {align*}
By the comparison rules, assuming that these limits exist. For simplicity, \begin {align*} L := \lim _{n \to \infty }\frac {x_{n+1}}{x_n}, M := \lim _{n \to \infty }\frac {y_{n+1}}{y_n}\\ L \leq M \end {align*}
Then, setting \(0\leq M <1\), by the ratio test, \(y_n \to 0\), because \(\{x_n\}\) is positive, we have, \(0 \leq L <1 \) as well, which again by the ratio test, implies \(x_n \to 0\) □
for sufficiently large \(n\).
We can partition the terms in the expression above into 3 categories:
Assuming \(n>a\), counting carefully, we would have have \(n-a\) terms of type (i), \(1\) term of type (ii), and \(n-(n-a)-1=a-1\) terms of type (iii). Now, we can consider the behaviour of the type (i) terms as the number of type (i) terms dominates as \(n\to \infty \). Now, we have: \begin {align*} \left |\frac {a^n}{n!}\right | = \left |\frac {a}{n}\right |\cdot \left |\frac {a}{n-1}\right | \cdot ... \cdot 1 \cdot \left | \frac {|a|^{|a|-1}}{(|a|-1)!}\right | \end {align*}
For, \(a \in \mathbb {R}-\mathbb {Z}\) and \(|a|>1\), we would have \(n - \lfloor |a| \rfloor \) terms of type (i), and \(\left \lfloor |a| \right \rfloor \) terms of type (iii), and no terms of type (ii) \begin {align*} \left |\frac {a^n}{n!}\right | = \left |\frac {a}{n}\right |\cdot \left |\frac {a}{n-1}\right | \cdot ... \cdot \left | \frac {|a|^{\lfloor |a| \rfloor }}{(\lfloor |a| \rfloor )!}\right | \end {align*}
Hence, for any \(|a|>1\) and \(a \in \mathbb {R}\), and WLOG, \(n>|a|\) the following inequality holds true: \begin {align*} \left |\frac {a^n}{n!}\right | < \left | \frac {a}{n} \right |\cdot \left | \frac {a}{a} \right |^{n -\lfloor |a| \rfloor -1} \cdot \left | \frac {|a|^{\lfloor |a| \rfloor }}{(\lfloor |a| \rfloor )!}\right | \end {align*}
This is because \(a< n-k\) for all terms of type (i), with this, we have: \begin {align*} \left |\frac {a^n}{n!}\right | < \left | \frac {a}{n} \right |\cdot 1^{n -\lfloor |a| \rfloor -1} \cdot \left | \frac {|a|^{\lfloor |a| \rfloor }}{(\lfloor |a| \rfloor )!}\right | = \frac {C}{n} \end {align*}
Where \(C = |a| \cdot \left | \frac {|a|^{\lfloor |a| \rfloor }}{(\lfloor |a| \rfloor )!}\right |\), hence the sequence is bounded by \(\frac {C}{n}\) when \(|a|>1\). For the case where \(|a|\leq 1\), for any \(n \in \mathbb {N}\) \begin {align*} \left |\frac {a^n}{n!}\right | \leq \frac {1^n}{n!} \leq \frac {1}{n} \end {align*}
Hence, in this case as well the sequence is bounded as well, with \(C=1\) as a potential value of the constant. Combining both cases, and the trivial condition of \(0\leq |a|\) we have: \begin {align*} 0 \leq \left |\frac {a^n}{n!}\right | \leq \frac {C}{n} \end {align*}
Because \(\frac {C}{n} \to 0\), by the squeeze theorem, \(\left |\frac {a^n}{n!}\right | \to 0\). Then, we can also conclude \(\frac {a^n}{n!} \to 0\).
(b) Defining the sequence \(b_n = \frac {a^n}{n!}\) for simplicity, then computing the ratio of
consecutive terms, we have: \begin {align*} \frac {b_{n+1}}{b_n} = \frac {a^{n+1}}{(n+1)!}\cdot \frac {n!}{a^n}= \frac {a}{(n+1)}\\ \lim _{n\to \infty } \frac {b_{n+1}}{b_n} = \lim _{n\to \infty } \frac {a}{(n+1)} = L =0 \end {align*}
Because \(0\leq L <1\), the sequence \(b_n\) converges to \(0\), by the sequence ratio test.
Hence, by the squeeze theorem, \(a_n^{\frac {1}{n}} \to 1\), as any \(a \in \mathbb {R}, a>0\) \(a^{\frac {1}{n}} \to 1\).
Hence, show that \(a_n \to 0\).
After, extensive manipulation, we get: \begin {align*} \lim _{n \to \infty } n \left [ \frac {a_{n+1}}{a_n} - \sqrt [3]{\frac {n}{n+1} } \right ] = -\frac {1}{6} \end {align*}
This implies \(\exists N \in \mathbb {N}\) such that \(\forall n > N\): \begin {align*} n \left [ \frac {a_{n+1}}{a_n} - \sqrt [3]{\frac {n}{n+1} } \right ] <0 \\ \implies \frac {a_{n+1}}{a_n} < \sqrt [3]{\frac {n}{n+1}} \\ \because \lim _{n \to \infty } \sqrt [3]{\frac {n}{n+1}} = 1 \implies \lim _{n \to \infty } \frac {a_{n+1}}{a_n} < \end {align*}