Real Analysis

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 \[ \overbrace{\underbrace{1 + \frac{1}{3} + \frac{1}{5} + \dots}_{> \pi} - \frac{1}{2} - \frac{1}{4} - \dots}^{< \pi} + \dots \]
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