11.8 Series

TutorialsWorshopsHomework 10.pdf

Analyze the following geometric series:

(a) \(\begin{aligned} \sum_{n=3}^{\infty} \frac{1}{3^n} \end{aligned}\): Easily, \(=\frac{1}{18}\)

(b) \(\begin{aligned} \sum_{n=2}^{\infty} \frac{(-5)^n}{8^{n+2}} \end{aligned}\): Easily, \(=\frac{25}{8^3\cdot13}\)

(c) \(\begin{aligned} \sum_{n=1}^{\infty} \frac{10^n + 5 \cdot 2^n}{6 \cdot 2^n} \end{aligned}\)
Since, we have a equivalence that: If \(\lim_{n\to\infty}\alpha^{n}\neq0\), then \(\sum\alpha_{n}\) diverges

Example: \(\sum_{n=1}^{\infty} \frac{n+1}{n+3}\), Since \(\lim_{n \to \infty} \frac{n+1}{n+3} = 1 \neq 0 \implies \sum \frac{n+1}{n+3} \text{ diverges}\)

However, the converse is not true: If \(\lim_{n \to \infty} a_n = 0 \not\Rightarrow \sum a_n \text{ is convergent}\)

Example: \(\sum\frac{1}{n}diverges,but\lim_{n\to\infty}\frac{1}{n}=0\)
Decide if the following series converges:

(a) \(\begin{aligned} \sum_{n=1}^{\infty} \frac{n^2}{n^2 - 2n} \end{aligned}\)

By theorem, it diverges since \(\lim_{n\to\infty}\frac{n^2}{n^2-2n}\neq0\)

(b) \(\begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n^2 + 2n} =\frac34 \end{aligned}\)

(c) \(\begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \end{aligned}\)

Since \(\begin{aligned}\sum_{n=1}^{\infty}\left|\frac{(-1)^{n}}{n^2}\right|=\end{aligned}\sum_{n=1}^{\infty}\frac{1}{n^2}\) is convergent, thus \(\begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \end{aligned}\) is convergent

(d) \(\begin{aligned} \sum_{n=1}^{\infty} \frac{\sin(n+3)}{n^2} \end{aligned}\)

Since \(\begin{aligned}\frac{\sin(n+3)}{n^2}\end{aligned}\leq\frac{1}{n^2}\) and \(\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{n^2}\end{aligned}\) is convergent, thus \(\begin{aligned} \sum_{n=1}^{\infty} \frac{\sin(n+3)}{n^2} \end{aligned}\) is convergent

(e) \(\begin{aligned} \sum_{n=1}^{\infty} \frac{1}{2^n + n} \end{aligned}\)

Since \(\frac{1}{2^{n}+n}\leq\frac{1}{2^{n}}\) and \(\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{2^{n}}\end{aligned}\) is convergent, thus \(\begin{aligned} \sum_{n=1}^{\infty} \frac{1}{2^n + n} \end{aligned}\) is convergent

(f) \(\begin{aligned} \sum_{n=1}^{\infty} \frac{n}{n^4 - 2} \end{aligned}\)

Since \(\begin{aligned}\frac{n}{n^4-2}\end{aligned}\leq\begin{aligned}\frac{n}{n^4-n}\end{aligned}\leq\frac{1}{n^3-1}\leq\frac{1}{n^2}\) and \(\begin{aligned}\sum_{n=1}^{\infty}\frac{1}{n^2}\end{aligned}\) is convergent, thus \(\begin{aligned} \sum_{n=1}^{\infty} \frac{n}{n^4 - 2} \end{aligned}\) is convergent

(g) \(\begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}} \end{aligned}\)

Since \(\begin{aligned}\left|\frac{(-1)^{n-1}}{\sqrt{n}}\right|\end{aligned}=\frac{1}{\sqrt{n}}\), but since \(\frac{1}{\sqrt{n}}>\frac{1}{n}>0\) and \(\sum\frac1n\) is divergent, thus \(\frac{1}{\sqrt{n}}\) is divergent

Thus we cannot use this method

Try to use D'Alembert Test (the Ratio Test):

For a series \(\sum a_n\), you compute the limit:

\(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)
- If \(L < 1\), the series converges absolutely.
- If \(L > 1\), the series diverges.
- If \(L = 1\), the test is inconclusive, and you need a different method to determine convergence.
no!

how to do?

Alternating Series Test: \(a_n = \frac{1}{\sqrt{n}}\).
1. Since \(\forall\varepsilon>0,\exists n_0=\frac{1}{\varepsilon^2},\forall n>n_0,\left|\frac{1}{\sqrt{n}}\right|=\frac{1}{\sqrt{n}}<\frac{1}{\sqrt{n_0}}=\varepsilon,\) hence \(\begin{aligned}\lim_{n\to\infty}\frac{1}{\sqrt{n}}=0.\end{aligned}\)
2. \(\begin{aligned}a_{n+1}=\frac{1}{\sqrt{n+1}}\quad\text{and}\quad a_{n}=\frac{1}{\sqrt{n}}.\end{aligned}Clearly:\begin{aligned}\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}\quad\text{for all}\quad n\geq1.\end{aligned}\)
By theorem, it is convergent

(h) \(\begin{aligned} \sum_{n=1}^{\infty} \frac{n!}{(-100)^n} \end{aligned}\)

We can see \(n!\) grow faster than \((-100)^n\), thus it is divergent

But we need to prove \(\frac{n!}{(-100)^{n}}\) become bigger and bigger

First we can take \(n\) is even to delete the minus signal, which is \(\frac{n!}{100^{n}}\)

We want \(\forall M,\forall n>n_0,\frac{n!}{100^{n}}>M\)

\(\frac{n!}{100^{n}}=\frac{1}{100}\cdot\frac{2}{100}\cdot\frac{3}{100}\cdot\ldots\cdot\frac{200}{100}\cdot\frac{201}{100}\cdot\frac{202}{100}\cdot\ldots\cdot\frac{n}{100}=\left(\frac{1}{100}\cdot\frac{2}{100}\cdot\frac{3}{100}\cdot\ldots\cdot\frac{200}{100}\right)\cdot\left(\frac{200}{100}\cdot\frac{201}{100}\cdot\frac{202}{100}\cdot\ldots\cdot\frac{n}{100}\right)>N\cdot2^{n-200}\)

Then we have \(N\cdot2^{n-200}>M\Rightarrow2^{n-200}>\frac{M}{N}\Rightarrow n_0=\log_2\left(\frac{M}{N}\right)+200\)

And we want \(n\) is even, then we can take \(n_0=2\left\lbrack\log_2\left(\frac{M}{N}\right)+200\right\rbrack\)

Finally, \(\forall n>n_0,\left|\frac{n!}{\left(-100\right)^{n}}\right|>M\)

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