Series

# Series

Given a sequence $$\{a_n\}$$, define $$\sum\limits_{i = p}^q a_i := a_p + a_{p + 1} + \dots + a_q$$.

Let $$s_n = \sum\limits_{k = 1}^n a_k$$, called the $$n$$th partial sum.

Then $$\{s_n\}$$ is a sequence, denoted $$\sum\limits_{n = 1}^\infty a_n$$ or sometimes $$\sum a_n$$, called an infinite series, which may or may not converge. If it does converge, say to some number $$s$$, then write $$\sum\limits_{n = 1}^\infty a_n = s$$.

Notice: $$\sum\limits_{n = 1}^\infty a_n = \lim\limits_{n \rightarrow \infty} s_n = \lim\limits_{n \rightarrow \infty} \sum\limits_{k = 1}^n a_k$$.

# Convergent Series

Theorem. A series $$\sum a_n$$ converges if and only if for every $$\epsilon > 0$$, there exists $$\mathcal{N}$$ such that $$m, n > \mathcal{N}$$ implies $$\left|\sum\limits_{k = n}^m a_k\right| < \epsilon$$.

Corollary. If $$\sum a_n$$ converges then $$\lim\limits_{n \rightarrow \infty} a_n = 0$$.

Theorem (non-negative series). If $$a_n \geq 0$$, then $$\sum a_n$$ converges if and only if the partial sums are bounded.

Proof. If $$a_n \geq 0$$, the partial sums are monotonically increasing. But they are also bounded. So they must converge. QED.

Theorem (comparison test).

1. If $$|a_n| \leq c_n$$ for $$n$$ large enough (say $$n > \mathcal{N}_1$$) and $$\sum c_n$$ converges then $$\sum a_n$$ converges.

2. If $$a_n \geq d_n \geq 0$$ for $$n$$ large enough (say $$n > \mathcal{N}_1$$) and $$\sum d_n$$ diverges then $$\sum a_n$$ diverges.

Proof of (a). Fix $$\epsilon > 0$$. Since $$\sum c_n$$ converges, there exists $$\mathcal{N}_2$$ such that $$m, n > \mathcal{N}_2$$ implies $$\left|\sum\limits_{k = n}^m c_k\right| < \epsilon$$. Let $$\mathcal{N} = \max\{\mathcal{N}_1, \mathcal{N}_2\}$$. For $$m, n > \mathcal{N}$$, $$\left|\sum\limits_{k = n}^m a_k\right| \leq \sum\limits_{k = n}^m |a_k| \leq \sum\limits_{k = n}^m c_k < \epsilon$$, as desired. QED.

Proof of (b). The contrapositive of (b) is if $$\sum a_n$$ converges, so does $$\sum d_n$$, which is shown in (a). QED.

Theorem (geometric series). If $$|x| < 1$$ then $$\sum\limits_{n = 0}^\infty x^n = \frac{1}{1 - x}$$. If $$|x| \geq 1$$ then $$\sum\limits_{n = 0}^\infty x^n$$ diverges.

Proof. Consider $$|x| < 1$$. Let $$s_n = 1 + x + \dots + x^n$$. Then $$s_n = \frac{1 - x^{n + 1}}{1 - x}$$. So, $$\lim\limits_{n \rightarrow \infty} s_n = \frac{1}{1 - x} \cdot \lim\limits_{n \rightarrow \infty} (1 - x^{n + 1}) = \frac{1}{1 - x}$$.

Consider $$|x| \geq 1$$. Terms don’t go to $$0$$ so the $$\sum x_n$$ diverges. QED.

Theorem (due to Cauchy). If $$a_1 \geq a_2 \geq \dots \geq 0$$ (monotonically decreasing) then $$\sum a_n$$ converges if and only if $$\sum 2^ka_{2^k}$$ converges.

Proof idea. Let $$s_n = a_1 + a_2 + \dots + a_n$$ and $$t_n = a_1 + 2a_2 + \dots + 2^na_{2^n}$$. So, $$s_n = (a_1) + (a_2 + a_3) + (a_4 + a_5 + a_6 + a_7) + \dots$$ and $$t_k = (a_1) + (a_2 + a_2) + (a_4 + a_4 + a_4 + a_4) + \dots$$. Notice if $$n < 2^k$$ then $$s_n < t_k$$. Consider $$2s_n = 2a_1 + 2a_2 + 2(a_3 + a_4) + 2(a_5 + a_6 + a_7 + a_8) + \dots + a_n$$ and $$t_k = a_1 + 2a_2 + 4a_4 + 8a_8 + \dots + 2^ka_{2^k}$$. Notice if $$n > 2k$$ then $$t_k < 2s_n$$.

Theorem. $$\sum\limits_{n = 1}^\infty \frac{1}{n^p}$$ converges if and only if $$p > 1$$.

Proof. When $$p \geq 0$$, the terms don’t go to $$0$$ so the series diverges. If $$p > 0$$ then $$\sum 2^k\frac{1}{2^{kp}} = \sum 2^{k(1-p)}$$, which is a geometric series, implying it converges if and only if $$2^{1 - p} < 1$$. $$2^{1 - p} < 1$$ if and only if $$1 - p < 0$$, as desired. QED.

For example,

1. $$\sum\limits_{n = 0}^\infty \frac{1}{n!}$$ converges to $$e$$.

Proof. Its partial sums are bounded by $$3$$, (see $$1 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$$). QED.

Its convergence is very rapid. Consider $$e - s_n = \frac{1}{(n + 1)!} + \frac{1}{(n + 2)!} + \dots < \frac{1}{(n + 1)!}\underbrace{\left(1 + \frac{1}{n + 1} + \frac{1}{(n + 1)^2} + \dots\right)}_{\text{goes to } \frac{n+1}{n}}$$, which goes to $$\frac{1}{n!n}$$.

$$e$$ is irrational. If $$e = \frac{m}{n}$$ then $$0 < \underbrace{n!(e - s_n)}_{\text{integer}} < \frac{1}{n}$$. There is no integer between $$0$$ and $$\frac{1}{n}$$!

Theorem (root test). Given $$\sum a_n$$, let $$\alpha = \limsup\limits_{n \rightarrow \infty} \sqrt[n]{|a_n|}$$. If $$\alpha < 1$$, the series converges. If $$\alpha > 1$$, the series divergers. If $$\alpha = 1$$, the test is inconclusive.

Proof (by comparison with the geometric series). If $$\alpha < 1$$, choose $$\beta$$ such that $$\alpha < \beta < 1$$. Then, by the definition of $$\limsup$$, there exists $$\mathcal{N}$$ such that $$n \geq \mathcal{N}$$ implies $$\sqrt[n]{|a_n|} < \beta$$. So, $$|a_n| < \beta^n|$$ for $$n \geq \mathcal{N}$$. But $$\sum \beta^n$$ converges, so $$\sum |a_n|$$ converges. Hence, $$\sum a_n$$ converges. If $$\alpha > 1$$, there exists a subsequence $$\sqrt[n_k]{|a_{n_k}|} \rightarrow \alpha > 1$$. So, $$|a_{n_k}| > 1$$ for infinitely many terms. Hence, terms do not go to zero. Thus, the series does not converge. If $$\alpha = 1$$, note that $$\sum \frac{1}{n}$$ diverges and $$\sum \frac{1}{n^2}$$ converges, both of which have $$\alpha = 1$$. QED.

Theorem (ratio test). $$\sum a_n$$ converges if $$\limsup\limits_{n \rightarrow \infty} \left|\frac{a_{n + 1}}{a_n}\right| < 1$$. The series diverges if $$\left|\frac{a_{n+1}}{a_n}\right| \geq 1$$ for $$n$$ large enough.

Proof (by comparison). If $$\limsup\limits_{n \rightarrow \infty} \left|\frac{a_{n + 1}}{a_n}\right| < 1$$, there exists $$\beta$$ such that $$\limsup\limits_{n \rightarrow \infty} \left|\frac{a_{n + 1}}{a_n}\right| < \beta < 1$$. Then, by definition of $$\limsup$$, there exists $$\mathcal{N}$$ such that $$\left|\frac{a_{n + 1}}{a_n}\right| < \beta < 1$$ for all $$n \geq \mathcal{N}$$. So, $$|a_{n + 1}| < \beta|a_n| < \beta^2|a_{n - 1}| < \beta^3|a_{n - 2}| < \dots$$. Hence, $$|a_{\mathcal{N} + k}| < \dots < \beta^k|a_\mathcal{N}|$$. Thus, $$\sum\limits_{k = 1}^n a_{\mathcal{N} + k} \leq a_\mathcal{N} \sum\limits_{k = 1}^n \beta^k$$. But $$\sum\limits_{k = 1}^n \beta^k$$ converges. Thus, $$\sum a_n$$ converges. For the divergence case, the terms don’t go to zero. QED.

For example,

1. If $$c_n$$ and $$z$$ are complex, $$\sum\limits_{n = 0}^\infty c_nz^n = c_0 + c_1z + c_2z^2 + \dots$$ is the power series.

Theorem. If $$\alpha = \limsup \sqrt[n]{|c_n|}$$ and $$R = \frac{1}{\alpha}$$ then $$\sum c_nz^n$$ converges if $$|z| < R$$ and it diverges if $$|z| > R$$. $$R$$ is called the radius of convergence.

Notice $$\sqrt[n]{|c_nz^n|} = z\sqrt[n]{|c_n|}$$, hence the radius.

Summation by Parts: Given sequences $$\{a_n\}$$ and $$\{b_n\}$$, let $$A_n = \sum\limits_{k = 0}^n a_k$$ for $$n > 0$$. Set $$A_{-1} = 0$$. Then $$\sum\limits_{n = p}^q a_nb_n = \sum\limits_{n = p}^{q - 1} A_n(b_n - b_{n + 1}) + A_qb_q - A_{p - 1}b_p$$.

Theorem. If $$A_n$$ is bounded and $$b_n$$ decreases and approaches $$0$$ then $$\sum a_nb_n$$ converges.

Proof. Say $$A_n \leq M$$. Given $$\epsilon > 0$$, there exists $$\mathcal{N}$$ such that $$b_\mathcal{N} \leq \frac{\epsilon}{2M}$$. For $$q \geq p \geq \mathcal{N}$$, $$\sum\limits_{n = p}^q a_nb_n = \sum\limits_{n = p}^{q - 1} A_n(b_n - b_{n + 1}) + A_qb_q - A_{p - 1}b_p \leq M\left|\sum\limits_{n = p}^{q - 1} (b_n - b_{n + 1}) + b_p + b_q\right| \leq 2Mb_p \leq 2Mb_\mathcal{N} < \epsilon$$. Thus, $$\sum a_nb_n$$ converges. QED.

Corollary. If $$|c_1| \geq |c_2| \geq \dots$$, $$c_i$$ are alternating in sign and they go to zero then $$c_n$$ converges (use $$a_n = (-1)^{n + 1}$$ and $$b = |c_n|$$).

# Absolute Convergence

Given series $$\sum a_n$$ and $$\sum b_n$$, let $$c_n = \sum\limits_{k = 0}^n a_kb_{n - k}$$. The product of the two series is defined as $$\sum c_n$$.

Theorem. If $$\sum a_n$$ converges to $$A$$ absolutely and $$\sum b_n$$ converges to $$B$$ absolutely then $$\sum c_n$$ converges to $$AB$$.

Say $$\sum a_n$$ converges alsolutely if $$\sum |a_n|$$ converges.

For example,

1. The series $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$$ converges, but not absolutely.

Theorem. If $$\sum a_n$$ converges absolutely then it converges.

# Rearrangements

Say $$\sum a_n = A$$. If the terms are rearranged, the series may not converge. If it does converge, it may not converge to $$A$$.

Theorem (due to Riemann). If $$\sum a_n$$ converges but not absolutely then a rearrangement can have any $$\limsup$$ and $$\liminf$$ you like. If it converges absolutely, all rearrangements have the same limits.

For example,

1. $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \ln 2$$. One rearrangement that converges to $$\pi$$ is $\overbrace{\underbrace{1 + \frac{1}{3} + \frac{1}{5} + \dots}_{> \pi} - \frac{1}{2} - \frac{1}{4} - \dots}^{< \pi} + \dots$