11.14 Series

Comparison test

Assume \(0 \leq a_n \leq b_n\)

a) If \(\sum b_n\) is convergent, then \(\sum a_n\) is convergent.

b) If \(\sum a_n\) diverges, then \(\sum b_n\) diverges.

Limit Comparison Criterion

Thm: If \(\lim_{n \to \infty} \frac{a_n}{b_n} = L \neq 0\)

Then \(\sum |a_n|\) converges \(\iff \sum |b_n|\) converges.

Equivalent: \(\sum |a_n|\) diverges \(\iff \sum |b_n|\) diverges.

Well Known Series

• Harmonic Series \(\sum \frac{1}{n}\)
• Geometric Series \(\sum a r^n\)
• Telescopic Series

• p-Series \(\sum\frac{1}{n^{p}}=\begin{cases}\text{Convergent} & \text{if }p>1\\ \text{Divergent} & \text{if }p\leq1\end{cases}\)​

• \(\sum_{n=1}^{\infty} \left(1 + \frac{1}{n^2}\right)^2\)

  Thm: If \(\sum a_n\) converges \(\Rightarrow \lim_{n \to \infty} a_n = 0\)

  Equiv. If \(\lim_{n \to \infty} a_n \neq 0 \Rightarrow \sum a_n\) diverges

  \(\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^2 = \left(\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)\right)^2 = 1^2 = 1 \neq 0 \Rightarrow \sum \left(1 + \frac{1}{n^2}\right)^2 \text{ diverges}\)

  Or using the Comparison Test

  \(a_n = 1\)

  \(b_n = \left(1 + \frac{1}{n^2}\right)^2 = 1 + \frac{2}{n^2} + \frac{1}{n^4}\)

  \(1 = a_n \leq b_n = \left(1 + \frac{1}{n^2}\right)^2\)

  \(\sum a_n = \sum 1 = 1 + 1 + 1 + \dots\) diverges.

• \(\sum_{n=1}^{\infty} \frac{n^2}{n^3 - 3}\)

  \(\lim_{n \to \infty} \frac{n^2}{n^3 - 3} = 0\)

  I'm not sure about the convergence of the series.

  For \(n \geq 2\): \(n^3 - 3 < n^3\) \(\Rightarrow \frac{1}{n^3 - 3} > \frac{1}{n^3}\)

  \(n^2>0,\forall n\), then \(\frac{n^2}{n^3-3}>\frac{n^2}{n^3}=\frac{1}{n}\)

  Let \(a_n = \frac{n^2}{n^3 - 3}\) and \(b_n = \frac{1}{n}\)

  \(\sum \frac{1}{n}\) is the harmonic series, which diverges.

  Using the Comparison Test \(\sum \frac{n^2}{n^3 - 3}\) diverges.

• \(\sum_{n=1}^{\infty} \frac{n - 1}{\sqrt{n^6 + 1}}\)

  \(\frac{n - 1}{\sqrt{n^6 + 1}} \approx \frac{n}{\sqrt{n^6}} = \frac{n}{n^3} = \frac{1}{n^2}\)

  We have \(\sum \frac{1}{n^2}\) converges

  \(\lim_{n \to \infty} \frac{\frac{n - 1}{\sqrt{n^6 + 1}}}{\frac{1}{n^2}} =\) \(\lim_{n \to \infty} \frac{(n - 1) \cdot n^2}{\sqrt{n^6 + 1}} =\) \(\lim_{n \to \infty} \frac{n^3 - n^2}{\sqrt{n^6 + 1}} =\) \(\lim_{n \to \infty} \frac{n^3\left(1 - \frac{1}{n}\right)}{n^3\sqrt{1 + \frac{1}{n^6}}} =\) \(\frac{1 - 0}{\sqrt{1 + 0}} = 1\)

  Thus it is convergent by Limit Comparison Criterion

• \(\sum_{n=1}^{\infty} \frac{\sqrt{2n^2 + 4n + 1}}{n^3 + 9}\)

  \(\frac{\sqrt{2n^2+4n+1}}{n^3+9}\approx\frac{n}{n^3}=\frac{1}{n^2}\)​

  \(\lim_{n\to\infty}\frac{n^2\sqrt{2n^2+4n+1}}{n^3+9}\) \(=\lim_{n\to\infty}\frac{n^3\sqrt{2+\frac{4}{n}+\frac{1}{n^2}}}{n^3\left(1+\frac{9}{n^3}\right)}\)​\(=\lim_{n\to\infty}\frac{\sqrt{2+\frac{4}{n}+\frac{1}{n^2}}}{\left(1+\frac{9}{n^3}\right)}\) \(=\sqrt2\)

  Thus convergent

• \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n + 1}}\)

  Since \(\sqrt{n} < \sqrt{n + 1}\) for all \(n\),

  \(\frac{1}{\sqrt{n + 1}} < \frac{1}{\sqrt{n}}\)

  Let \(a_n = \frac{1}{\sqrt{n + 1}}\) and \(b_n = \frac{1}{\sqrt{n}}\).

  \(\lim_{n\to\infty}\frac{\frac{1}{\sqrt{n+1}}}{\frac{1}{\sqrt{n}}}=\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n+1}}=\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n+1}}\cdot\frac{\sqrt{n}-1}{\sqrt{n}-1}=\lim_{n\to\infty}\frac{n-\sqrt{n}}{n-1}=\lim_{n\to\infty}\frac{1-\frac{1}{\sqrt{n}}}{1-\frac{1}{n}}=1\)

  Since \(b_n = \frac{1}{\sqrt{n}}\) is divergent, then \(a_n = \frac{1}{\sqrt{n + 1}}\) is divergent.

• \(\sum_{n=1}^{\infty} \frac{1}{n!}\) with \(a_{n+1} = \frac{1}{(n+1)!}\)

  \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} = \lim_{n \to \infty} \frac{n!}{(n+1)!}\)

  \(= \lim_{n \to \infty} \frac{1}{n+1} = 0 < 1\)

  Thus, \(\sum \frac{1}{n!}\) converges absolutely.

• \(\sum_{n=1}^{\infty} \frac{n!}{n^n}\)

  \(\lim_{n \to \infty} \frac{n!}{n^n} = 0\)

  I'm sure about the convergence of the series.

  \(a_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}}\)

  From previous theorem: \(\sum_{n=1}^{\infty} \frac{n!}{n^n}\)

  \(\lim_{n\to\infty}\left|\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^{n}}}\right|=\lim_{n\to\infty}\frac{(n+1)!\cdot n^{n}}{(n+1)^{n+1}\cdot n!}=\lim_{n\to\infty}\frac{(n+1)\cdot n^{n}}{(n+1)^{n+1}}=\lim_{n\to\infty}\frac{n^{n}}{(n+1)^{n}}\)​

  \(=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{n}=\lim_{n\to\infty}\left(\frac{1}{1+\frac{1}{n}}\right)^{n}=\frac{1}{e}<1\)

  Thus convergent

• \(\sum_{n=1}^{\infty} \left( \frac{5 - 3n^3}{7n^3 + 2} \right)^n\)

  \(\lim_{n\to\infty}\sqrt[n]{\left|\left(\frac{5-3n^3}{7n^3+2}\right)^{n}\right|}=\lim_{n\to\infty}\left|\frac{5-3n^3}{7n^3+2}\right|=\left|\lim_{n\to\infty}\frac{5-3n^3}{7n^3+2}\right|=\frac37<1\)

  Since \(\frac{3}{7} < 1\), the series \(\sum \left( \frac{5 - 3n^3}{7n^3 + 2} \right)^n\) absolutely converges.

  The it is convergent

• A series is conditionally convergent if \(\sum a_n\) converges and \(\sum |a_n|\) diverges.

• Example: \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\), we will prove that this series converges.

  \(\sum \left|\frac{(-1)^n}{n}\right| = \sum \frac{1}{n}\) diverges.

  Partial Sums:
  ​\(\lim_{n \to \infty} S_n = \sum_{n=1}^{\infty} S_n\)

  \(\begin{aligned}S_1 & =-1\\ S_2 & =-1+\frac12\\ S_3 & =-1+\frac12-\frac13\\ S_4 & =-1+\frac12-\frac13+\frac14\end{aligned}\)​

  Claim 1: \(S_1 < S_3 < \dots < S_{2n+1} < \dots\)

  \(\begin{aligned}S_{2n+1} =-1+\frac12-\cdots+\frac{1}{2n}-\frac{1}{2n+1}\\S_{2n+3} =-1+\frac12-\cdots+\frac{1}{2n+2}-\frac{1}{2n+3}\\ S_{2n+3}-S_{2n+1} =\frac{1}{2n+2}-\frac{1}{2n+3}>0\end{aligned}\)​

  Then \(S_{2n+3} - S_{2n+1} > 0 \implies S_{2n+3} > S_{2n+1} \quad \forall n\)

  \((S_{2n+1})\) is increasing.

  Claim 2: \(\dots < S_{2n} < S_{2n-2} < \dots < S_2\)

  Proof: Exercise

  Claim 3: \(S_{2n+1} < S_{2n+2}\)

  Proof: Exercise

  Then

  \(S_1 < S_3 < \dots < S_{2n+1} < \dots < S_2\) Claim 3

  \(S_{2n+1}\) is increasing and has an upper bound \(S_2\)

  \(\implies \exists \lim_{n \to \infty} S_{2n+1} = L_1\)

  \(S_1 < \dots < S_{2n} < S_{2n-2} < \dots < S_2\) Claim 3

  \(S_{2n}\) is decreasing and has a lower bound \(S_1\)

  By monotonic theorem, \(S_{2n}\) converges.

  \(\lim_{n \to \infty} S_{2n} = L_2\)

  \(S_{2n+1} - S_{2n} = \frac{-1}{2n+1}\)

  \(\lim_{n \to \infty} (S_{2n+1} - S_{2n}) = \lim_{n \to \infty} \frac{-1}{2n+1} = 0\)

  \(L_1 - L_2 = 0\)

  \(\implies L_1 = L_2\)

  \(\therefore \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\) converges

• Find a \(t\) such that the following series converges:

  \(\begin{aligned}\sum_{n=1}^{\infty}n^2t^{n}\end{aligned}\)​

  \(\lim_{n \to \infty} \left| \frac{(n+1)^2 t^{n+1}}{n^2 t^n} \right| = \lim_{n \to \infty} \left( \frac{(n+1)^2}{n^2} \right) |t|\)

  \(= \lim_{n \to \infty} \left( \frac{n+1}{n} \right)^2 |t|\)

  \(=|t|\cdot\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^2=|t|<1\), thus \(-1<t<1\)​

• \(\sum_{n=1}^{\infty} \frac{t^n}{2^n} = \sum_{n=1}^{\infty} \left( \frac{t}{2} \right)^n\)

  \(= \frac{t}{2} + \frac{t^2}{2^2} + \frac{t^3}{2^3} + \dots\)

  \(= \sum_{n=0}^{\infty} \frac{t}{2} \left( \frac{t}{2} \right)^n\)

  \(|r| = \left| \frac{t}{2} \right| < 1 \implies \frac{|t|}{2} < 1 \implies |t| < 2\)

  Thus \(-2<t<2\)

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