KindiakMath

A Handful of Convergence Tests

Now we begin on the slew of convergence tests that will be essential in our study in real analysis. Let’s first recall our all-important comparison test.

Theorem 1 (Comparison Test). Let \{x_n\}, \{y_n\} be non-negative sequences with partial sums \{s_n\}, \{t_n\}.

Suppose for any n, 0 \leq x_n \leq y_n.

  • If t_\infty converges, then s_\infty converges.
  • If s_\infty diverges, then t_\infty diverges, and s_\infty = t_\infty = \infty.

Closely related to the comparison test is the limit comparison test. This provides a more direct computation tool to help us establish convergence.

Theorem 2 (Limit Comparison Test). Let \{x_n\}, \{y_n\} be non-negative sequences with partial sums \{s_n\}, \{t_n\}. Suppose y_n \neq 0 for any n.

Suppose there exists \rho \in \mathbb R such that x_n/y_n \to \rho.

  • Suppose \rho > 0. Then either both s_\infty and t_\infty converge or both s_\infty and t_\infty diverge.
  • Suppose \rho = 0. Then t_\infty converges implies s_\infty converges.

Proof. The key insight is to apply the comparison test. Since x_n/y_n \to \rho > 0, for \epsilon, there exists N \in \mathbb N such that

\displaystyle n > N \quad \Rightarrow \quad \left| \frac{x_n}{y_n} - \rho \right| < \epsilon \quad \Rightarrow \quad (\rho -\epsilon) y_n < x_n < (\rho +\epsilon) y_n.

Suppose \rho > 0. Then choosing \epsilon = \rho/2 > 0 yields

\displaystyle \frac{\rho}{2} \cdot y_n < x_n < \frac{\rho}{2} \cdot y_n.

Convergence properties follows from the comparison test.

Suppose \rho = 0. Then choosing \epsilon = 1 > 0 yields

-y_n < x_n < y_n.

Since x_n \geq 0, 0 \leq x_n \leq y_n. Convergence properties then follows from the comparison test.

Another common convergence test is the ratio test, whose result infamously appeared in the 2017 H2 Math examination in Singapore.

Theorem 3 (Ratio Test). Let \{x_n\} be a non-negative sequence with partial sum \{s_n\}. Suppose there exists \rho \in \mathbb R such that x_{n+1}/x_n \to \rho \geq 0.

  • If \rho < 1, then s_\infty converges.
  • If \rho > 1, then s_\infty diverges.
  • If \rho = 1, then the test is inconclusive.

Proof. The criterion looks awfully similar to that of the limit comparison test, but clearly the results are rather different. Heuristically, for large n, we have

x_n \approx x_{n-1} \rho \approx \cdots \approx x_N \rho^{n-N},

which looks an awful lot like our geometric series. That turns out to be our weapon of choice here. Fix \epsilon > 0 which we will judicially choose for our proof later on. Since x_{n+1} / x_n \to \rho, there exists N \in \mathbb N such that

\displaystyle n > N \quad \Rightarrow \quad \left| \frac{ x_{n+1} }{ x_n } - \rho \right| < \epsilon\quad \Rightarrow \quad (\rho - \epsilon) x_n < x_{n+1} < (\rho + \epsilon) x_n.

If \rho = 0, choosing \epsilon = 1/2 =: r \in (0, 1) yields

\displaystyle x_{n+1} < r x_n < \cdots < r^{n-N+1} x_N.

Summing both sides,

\displaystyle \sum_{k=1}^\infty x_{k+1} \leq \sum_{k=1}^\infty x_{k+1} < x_N r^{2-N} \cdot \sum_{k=1}^{\infty} r^{k-1} < \infty.

If 0 < \rho < 1, choosing \epsilon := (1-\rho)/2 > 0 yields r := \rho + \epsilon \in (0, 1) so that

\displaystyle x_{n+1} < r x_n < \cdots < r^{n-N+1} x_N,

and a similar result follows.

If \rho > 1, choosing \epsilon := (\rho - 1)/2 > 0 yields r := \rho - \epsilon > 1 so that

\displaystyle x_{n+1} > r^{n-N+1} x_N.

Since the geometric series diverges when r > 1, so does s_\infty.

What’s with the “inconclusive” bit? Well, the sequence defined by x_n = 1 satisfies x_{n+1}/x_n = 1 \to 1, and in this case s_\infty clearly diverges.

However, now consider the sequence defined by

\displaystyle x_n = \frac 1n - \frac 1{n+1} = \frac 1{n(n+1)}.

Here,

\displaystyle \begin{aligned} \frac{x_{n+1}}{x_n} &= \frac{1}{(n+1)(n+2)} /\frac{1}{n(n+1)} \\ &= \frac{n(n+1)}{(n+1)(n+2)} \\ &= \frac{n}{n+2} = \frac 1{1+2/n} \to \frac 1{1+0} = 1. \end{aligned}

However,

\displaystyle \begin{aligned} s_n &= \sum_{k=1}^n \left( \frac{1}{k} -\frac 1{k+1} \right) \\ &= \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac 1{k+1} \\ &= 1 + \sum_{k=2}^n \frac 1k - \sum_{k=2}^n \frac 1k - \frac{1}{n+1} \\ &= 1 - \frac{1}{n+1} \to 1 - 0 = 1. \end{aligned}

Thus, s_\infty converges. This means that knowing x_{n+1} /x_n \to 1 tells us nothing about the convergence of s_\infty.

Another perspective on the ratio test is the root test, which considers not the quantity x_{n+1}/x_n \to \rho, but the quantity x_n^{1/n} \to \rho.

Theorem 4 (Root Test). Let \{x_n\} be a non-negative sequence with partial sum \{s_n\}. Suppose there exists \rho \in \mathbb R such that x_n^{1/n} \to \rho \geq 0.

  • If \rho < 1, then s_\infty converges.
  • If \rho > 1, then s_\infty diverges.
  • If \rho = 1, then the test is inconclusive.

Proof. Fix \epsilon \in (0,\rho) which we will judicially choose for our proof later on. Since x_n^{1/n} \to \rho, there exists N \in \mathbb N such that

n > N \quad \Rightarrow \quad |x_n^{1/n} - \rho| < \epsilon \quad \Rightarrow \quad (\rho-\epsilon)^n < x_n < (\rho+\epsilon)^n.

If \rho < 1, then choosing \epsilon = (1-\rho)/2 > 0 yields r := \rho + \epsilon \in (0, 1) so that x_n < r^n < r_N \cdot r^{n-N}. Since the geometric series converges, by the comparison test, s_\infty converges.

If \rho > 1, then choosing \epsilon = (\rho-1)/2 > 0 yields r := \rho + \epsilon > 1 so that r^n > r_N \cdot r^{n-N}. Since the geometric series diverges, by the comparison test, s_\infty converges.

For the inconclusiveness of \rho = 1, we can verify the conditions using the same counterexamples in the ratio test inconclusiveness.

It turns out that the root test is stronger than the ratio test in the following sense.

Theorem 5. Let \{x_n\} be a non-negative sequence. Suppose there exists \rho \in \mathbb R such that x_{n+1}/x_n \to \rho. Then x_n^{1/n} \to \rho.

Proof. Fix \epsilon > 0. Suppose \rho > 0 so that x_n > 0. Since x_{n+1}/x_n \to \rho, for any k_1 \in (0, \rho), there exists N_1 \in \mathbb N such that

\displaystyle n > N_1 \quad \Rightarrow \quad \left| \frac{x_{n+1}}{x_n} - \rho \right|< k_1 \quad \Rightarrow \quad (\rho - k_1)x_n < x_{n+1} < ( \rho + k_1)x_n.

Similar to the previous proofs, for n > N_1+1,

\displaystyle (\rho - k_1)^{n-{N_1}} x_{N_1} < x_n < ( \rho + k_1)^{n-{N_1}} x_{N_1}.

Doing a bit of algebra,

\displaystyle \frac{ (\rho - k_1)^n }{ \rho^n } \cdot \frac{ x_{N_1} }{ (\rho - k_1)^{N_1} } < \frac{ x_n }{ \rho^n } < \frac{ (\rho + k_1)^n }{ \rho^n } \cdot \frac{ x_{N_1} }{ (\rho + k_1)^{N_1} }.

Taking n-th roots and doing a bit more algebra,

\displaystyle (\rho - k_1) \cdot a_-^{1/n} - \rho < x_n^{1/n} - \rho < (\rho + k_1)\cdot a_+^{1/n} - \rho,

where a_{\pm} = x_{N_1}/ (\rho \pm k_1)^{N_1} > 0. Since a_{\pm}^{1/n} \to 1, for any k_2 > 0, there exists N_2 \in \mathbb N such that

n > N_2 \quad \Rightarrow \quad |a_{\pm} - 1| < k_2.

Therefore, for n > \max\{N_1, N_2\}, we have

\displaystyle (\rho - k_1) (1 - k_2) - \rho < x_n^{1/n} - \rho < (\rho + k_1) (1 + k_2) - \rho,

which simplifies to

\displaystyle - (k_1 + k_2 (\rho-k_1)) < x_n^{1/n} - \rho < k_1 + k_2 (\rho + k_1).

By the triangle inequality,

|x_n^{1/n} - \rho| < k_1 + k_2 \rho + k_1 k_2.

In our arguments, all we needed to do is choose k_1 and k_2. To that end, choose k_1 = \max\{ \epsilon/3, \rho/2 \} and k_2 = \min\{ \epsilon/\rho, \epsilon, 3\}/3.

The case \rho = 0 is left as an exercise in rigorously implementing the following big-picture idea: x_{n+1}/x_n \approx 0 implies x_n \approx 0 so that x_n^{1/n} \approx 0.

These tests constitute the most common convergence tests. There are many more variants, even methods to relax the limiting conditions, but these convergence tests provide some of the most versatile tools to verify (or disprove, even) the convergence of various series.

—Joel Kindiak, 28 Dec 24, 1436H

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