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Some convergence tests for sequences

Read Section 2.2. Also, review the materials about sequences from Calculus II.

Check your understanding against the following questions:

• For $c > 0$, what is $\lim c^n$? (there are two cases to consider).
• What is the ratio test for sequences?

Let $P$ be a property of the real numbers (e.g. nonnegative), we say that a sequence of real numbers is $P$ if every term of the sequence has that property.

Proposition 1. The sequence $(c^n)$ either converges to $0$ if $0 < c <1$ or diverges to $+\infty$ if $c > 1$.

Proof. First, suppose $0 < c < 1$. Then $$ c > c^2 > c^3 > \cdots $$ So, $(c^n)$ is convergent because it is a decreasing sequence of real numbers bounded below by $0$ (Week 5 Theorem 3 Monotone Convergence Theorem (MCT)).

Let $L$ be the limit. On the one hand, the sequence $c(c^n) = (c^{n+1})$ has limit $cL$, on the other hand, it is simply $(c^n)$ with the first term dropped, so it must have the same limit $L$. Thus, by the uniqueness of limit, $cL =L$. That is $L(1-c) = 0$ and since $1-c > 0$ is positive (nonzero), $L$ must be $0$.

Now if $c > 1$, then $0 < 1/c < 1$. And so, $1/c^n \to 0$. Consequently, for any $B > 0$, there is an $N$ so that $1/c^n < 1/B$, i.e. $B < c^n$ for all $n \ge N$. This shows that $(c^n)$ diverges to $+\infty$.

Proposition 2. For $c >0$, $c^{1/n} \to 1$.

Proof. First suppose $c \ge 1$. Then

$$c \ge c^{1/2} \ge c^{1/3} \cdots \ge 1$$

So, $(c^{1/n})$ is convergent by the MCT. Since each term of the sequence is at least $1$, so is its limit $L$. Any subsequence, in particular the subsequence $(c^{1/2n})$, of $(c^{1/n})$ converges to $L$. But then $c^{1/n} = (c^{1/2n})^2 \to L^2$. So, $L^2 = L$ by uniqueness of limits. Canceling $L$ on both sides, we get $L =1$.

If $0< c \le 1$ then $1/c \ge 1$ and so $1/c^{1/n} = (1/c)^{1/n} \to 1$ by the above argument. Taking reciporcial, we get $c^{1/n} \to 1/1 = 1$.

Proposition 3. (Ratio test) Let $(x_n)$ be a sequence of nonzero numbers and $L:=\lim |x_{n+1}|/|x_n|$ exists. Then $(x_n)$ is null if $L < 1$ or unbounded if $L > 1$.

If $L =1$, the test is inconclusive. (consider the two sequences $(1/n)$, and $((-1)^n)$

Proof. Suppose $L < 1$. Since $\left(|x_{n+1}|/|x_n|\right)$ is a convergent sequence of non-negative numbers, its limit $L$ is nonnegative as well (Week 5 Proposition 8). Thus $0 \le L < c:=\dfrac{1+L}{2} <1$ and there exists $N$ such that whenever $n \ge N$, $|x_{n+1}|/|x_n| < c$ or equivalently $|x_{n+1}| < c|x_n|$. As a result, for any $k \in \Nn$, $$0 \le |x_{N+k}| < c|x_{N+k-1}| < \cdots < c^{k}|x_{N}|.$$ Since $0 \le c < 1$, $(c^k|x_N|)$ is null and hence it follows from the above inequalities that $(x_n)$ is null. The case $L >1$ can be proved similarly, left as exercise.

For some interesting applications of the ratio test, read Example 2.2.13 and 2.2.14

The limits of some sequences (optional)

1. $\ds \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$ exists.

Proof. We contend that $a_n := \left(1+ \frac{1}{n}\right)^n$ ($n \in \Nn$) form an increasing sequence that is bounded above.

It follows from the Binomial theorem that $$ a_n = \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}. $$ Since $\ds \binom{n}{k}\frac{1}{n^k} = \frac{n(n-1)\cdots (n-k+1)}{k!}\frac{1}{n^k} < \frac{1}{k!} \le \frac{1}{2^{k-1}}$ for all $k \ge 1$, $\ds a_n \le 1 + \sum_{k=1}^n \frac{1}{2^{k-1}} = 1 + 2 = 3$.

Clearly for each $0 \le k \le n$, $$ \binom{n+1}{k}\frac{1}{(n+1)^k} = \frac{1}{k!}\prod_{i=0}^k \left(1 - \frac{i}{n+1}\right) \ge \frac{1}{k!}\prod_{i=0}^k \left(1 - \frac{i}{n}\right) = \binom{n}{k}\frac{1}{n^k}. $$ So, $$ a_n = \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k} \le \sum_{k=0}^n \binom{n+1}{k}\frac{1}{(n+1)^k} < \sum_{k=0}^{n+1} \binom{n+1}{k}\frac{1}{(n+1)^k} = a_{n+1} $$

and this is the contention.

The limit above is approximately $2.718281828$ and is denoted by $e$. This definition of $e$ is elementary. However, in developing properties of the exponential function, it is often more flexible to use the following definition.

We first define $\ln(x)$ as $\ds \int_1^x \frac{1}{t} dt$. Note that $\ln(x)$ is a strictly increasing function (from $[1,\infty)$ to $[0,\infty)$) and hence its has an inverse. We denote it by $\exp(x)$. Let $e$ be $\exp(1)$, the value of $\exp(x)$ at $1$. In other words, $e$ is the real number so that the area under the curve $y=1/x$ from $1$ to $e$ is $1$. We extend the domain of $\exp(x)$ to $\Rr$ by defining $\ds e^x = \frac{1}{e^{-x}}$ for $x < 0$.

As an application, let us show that for any $c >0$, $\ds f(x) = \left(1+\frac{c}{x}\right)^x$ is increasing. In particular, (when $c=1$) this provides another proof that the sequence $\ds \left(\left(1+\frac{1}{n}\right)^n \right)$ is increasing.

Since $\exp(x)$ is increasing, it sufficies to show that $\ds \ln f(x) = x\ln\left(1+ \frac{c}{x}\right)$ is increasing or equivalently that its derivative $\ds \ln \left(1 + \frac{c}{x}\right) - \frac{c}{x+c}$ is strictly positive. By recognizing $\ds \ln\left(1 + \frac{c}{x}\right)$ as the area represented by $\ds \int_x^{x+c} \frac{1}{t} dt$ and that $1/x$ is decreasing, the inequalities follow immediately:

$$ \frac{c}{x+c} < \ln\left(1 + \frac{c}{x}\right) < \frac{c}{x}. $$

As a bouns, by multiplying $x$ throughout, we get

$$ \frac{cx}{x+c} < x\ln\left(1 + \frac{c}{x}\right) < c. $$ Letting $x \to \infty$, we have $\ds \lim_{x\to \infty}x\ln\left(1 + \frac{c}{x}\right) = \lim_{x\to \infty} \ln \left(1 + \frac{c}{x}\right)^x = c$ by the squeeze lemma. Consequently, we show that $\ds \lim_{x\to \infty} \left(1 + \frac{c}{x}\right)^x = e^c$ for $c >0$. The case $c < 0$ is left as an exercise.

2. $n^a \ll b^n \ll n! \ll n^n$ ($a \ge 0$ and $b > 1$). That is the ratio of one over the next tends to $0$ as $n \to \infty$.

Roughly speaking, that means any exponential function dominates any powers, factorial dominates any exponential function, and $n^n$ dominates factorial for all sufficiently large $n$.

Proof. $\ds \frac{(n+1)^a}{b^{n+1}}\frac{b^n}{n^a} = \left(1+\frac{1}{n}\right)^a \frac{1}{b} \to \frac{1}{b} < 1$, so $\ds \frac{n^a}{b^n} \to 0$ by the ratio test.

Also $\ds \frac{b^{n+1}}{(n+1)!}\frac{n!}{b^n} = \frac{b}{n+1} \to 0$, thus $\ds \frac{b^n}{n!} \to 0$ by the ratio test.

Finally, $\ds 0 \le \frac{n!}{n^n} = 1 \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots \frac{1}{n} < \frac{1}{n}$. Thus, $\dfrac{n!}{n^n} \to 0$ by the squeeze lemma.

3. $\lim n^{1/n} = 1$.

Proof. Since $1 \le n^{1/n}$ for all $n$, it suffices to so that for any $\varepsilon >0$, $1 \le n^{1/n} < 1+\varepsilon$ for all $n$ sufficiently large. But this follows immediately from $\dfrac{n}{(1+\varepsilon)^n} \to 0$ (any power is domainated by any exponential function).

Divergence of sequences involving basic trigonmetric functions

The sequence $(\tan(n))_n$ is divergent.

Proof. Suppose $\tan(n) \to L$ for some $L \in \Rr$ then so does $\tan(n+1)$.

Since $\pi$ is irrational, so $\tan(n+1)$$ is defined and $$\tan(n+1) = \dfrac{\tan(n) + \tan(1)}{1-\tan(n)\tan(1)}.$$

Thus, $\tan(n+1)(1-\tan(n)\tan(1)) = \tan(n) + \tan(1)$ for all $n$.

Letting $n \to \infty$, we get $L(1-L\tan(1)) = L+\tan(1)$. Since $\tan(1) \neq 0$, we get $L^2 =-1$ a contradiction. Therefore, $(\tan(n))$ diverges.

The sequence $(\sin(n))_n$ is divergent.

Proof. Suppose on the contrary that $\sin(n) \to L$. Then $\sin(n+1)$ and $\sin(n-1)$ also tends to $L$ as $n \to \infty$. Consider the identities, \begin{align*} \sin(n+1) - \sin(n-1) &\equiv 2\cos(n)\sin(1) \\ \sin(n+1) + \sin(n-1) &\equiv 2\sin(n)\cos(1) \\ \sin^2(n) +\cos^2(n) &\equiv 1 \end{align*} By letting $n \to \infty$, it follows from the 1st one that $\cos(n) \to 0$ and from 2nd one that that $L = 0$. And consequently, from the 3rd identity that $0 = 1$ which is absurd.

Exercise. Show that $(\cos(n))_n$ is divergent.

Remark. In fact, something much stronger is true: For any $\theta \in [-1,1]$, there is a subsequence of $(\sin(n))$ that converges to $\theta$.

One can deduce that from Kronecker's theorem in Diophantine Approximation. More generally, it is a baby example of a sequence that is uniformly distributed mod 1.