11.15 Series
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Show that if \(a_n > 0\) and \(\lim_{n \to \infty} n a_n = l\) with \(l \neq 0\), then the series \(\sum a_n\) diverges.
Consider \(\lim_{n\to\infty}na_{n}=l=\lim_{n\to\infty}\frac{a_{n}}{\frac{1}{n}}=\frac{\lim_{n\to\infty}a_{n}}{\lim_{n\to\infty}\frac{1}{n}}\), then since \(\sum\frac{1}{n}\) is divergent.
Then \(\sum a_n\) is also divergent
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Assume \(a_n > 0\) and \(\lim n^2 a_n\) exists. Show that \(\sum a_n\) converges.
Consider \(\lim_{n\to\infty}n^2a_{n}=l=\lim_{n\to\infty}\frac{a_{n}}{\frac{1}{n^2}}=\frac{\lim_{n\to\infty}a_{n}}{\lim_{n\to\infty}\frac{1}{n^2}}\), then since \(\sum\frac{1}{n^2}\) is convergent.
Then \(\sum a_n\) is also convergent
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Decide if the following series converge or diverge:
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\(\sum_{n=1}^{\infty} \frac{16 + (-2)^n}{n 2^n}\)
Firstly, \(\sum_{n=1}^{\infty}\frac{16+(-2)^{n}}{n2^{n}}=\sum_{n=1}^{\infty}\left(\frac{16}{n2^{n}}+\frac{(-2)^{n}}{n2^{n}}\right)=\sum_{n=1}^{\infty}\left(\frac{1}{n2^{n-4}}+\frac{1}{n}\cdot\left(-1\right)^{n}\right)=\sum_{n=1}^{\infty}\frac{1}{n2^{n-4}}+\sum_{n=1}^{\infty}\frac{1}{n}\cdot\left(-1\right)^{n}\)
Thus let analysis \(\sum_{n=1}^{\infty}\frac{1}{n2^{n-4}}\) and \(\sum_{n=1}^{\infty}\frac{1}{n}\cdot\left(-1\right)^{n}\)
Since \(\frac{1}{n2^{n-4}}\leq\frac{1}{2^{n-4}}\) and \(\sum\frac{1}{2^{n-4}}\) is convergent, thus \(\sum_{n=1}^{\infty}\frac{1}{n2^{n-4}}\) is convergent
By this, we know \(\sum_{n=1}^{\infty}\frac{1}{n}\cdot\left(-1\right)^{n}\) is conditionally convergent
Thus by theorem, \(\sum_{n=1}^{\infty} \frac{16 + (-2)^n}{n 2^n}\) is convergent 2. \(\sum_{n=1}^{\infty} \frac{1}{(n^2 + n)^q}\)
Let \(\frac{1}{(n^2+n)^{q}}\leq\frac{1}{n^{2p}}\)
If \(2p>1\Rightarrow p>\frac12\), \(\sum\frac{1}{n^{2p}}\) is convergent, thus \(\sum_{n=1}^{\infty} \frac{1}{(n^2 + n)^q}\) is convergent
Let \(\frac{1}{2n^{2p}}\leq\frac{1}{(n^2+n)^{q}}\)
If \(2p\leq1\Rightarrow p \leq\frac12\), \(\sum\frac{1}{2n^{2p}}\) is divergent, thus \(\sum_{n=1}^{\infty} \frac{1}{(n^2 + n)^q}\) is divergent 3. \(\sum_{n=1}^{\infty} \frac{\sqrt{n^2 - 1}}{\sqrt{n^5 + 1}}\)
\(\frac{\sqrt{n^2-1}}{\sqrt{n^5+1}}=\sqrt{\frac{n^2-1}{n^5+1}}\leq\frac{n^2-1}{n^5+1}\leq\frac{n^3}{n^5+1}\leq\frac{n^3}{n^5}=\frac{1}{n^2}\)
Obviously, it is convergent 4. \(\sum_{n=1}^{\infty} \left( 1 - \frac{1}{n} \right)^{n^2}\)
Since \(\left(1-\frac{1}{n}\right)^{n^2}=\left(\left(1-\frac{1}{n}\right)^{n}\right)^2=\left(\left(\frac{1}{1+\frac{1}{n}}\right)^{n}\right)^2\), then \(\lim\left(1-\frac{1}{n}\right)^{n^2}=\lim\left(\left(\frac{1}{1+\frac{1}{n}}\right)^{n}\right)^2=\frac{1}{e^2}\neq0\)
Thus it is not convergent by theorem
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Prove that \(\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\) converges conditionally.
Alternating Series Test: \(a_n = \frac{1}{\sqrt{n}}\).
- 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 \(\begin{aligned}\lim_{n\to\infty}\frac{1}{\sqrt{n}}=0.\end{aligned}\)
- \(\begin{aligned}a_{n+1}=\frac{1}{\sqrt{n+1}}\quad\text{and}\quad a_{n}=\frac{1}{\sqrt{n}}.\end{aligned}Clearly:\begin{aligned}\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}\quad\text{for all}\quad n\geq1.\end{aligned}\)
By theorem, it is convergent
Or
Consider \(\sum=-1+\frac{1}{\sqrt2}-\frac{1}{\sqrt3}+\frac{1}{\sqrt4}+\cdots\)
Let use partial sum, consider a new sequence \(b_n\)
\(b_1=-1+\frac{1}{\sqrt2},b_2=-\frac{1}{\sqrt3}+\frac{1}{\sqrt4},\ldots,b_{n}=-\frac{1}{\sqrt{2n+1}}+\frac{1}{\sqrt{2n+3}}\)
Then \(\sum a_n=\sum b_n\)
Since \(b_{n}=-\frac{1}{\sqrt{2n+1}}+\frac{1}{\sqrt{2n+3}}=\frac{\sqrt{2n+1}-\sqrt{2n+3}}{\sqrt{\left(2n+1\right)\left(2n+3\right)}}=\frac{1}{(\sqrt{2n+2}+\sqrt{2n+1})\cdot\sqrt{2n+1}\cdot\sqrt{2n+2}}\)
\(b_{n}=\frac{1}{(\sqrt{2n+2}+\sqrt{2n+1})\cdot\sqrt{2n+1}\cdot\sqrt{2n+2}}\leq\frac{1}{2\sqrt{2n}\cdot2n}=\frac{1}{4\sqrt2n^{\frac32}}\)
Thus since \(\sum\frac{1}{4\sqrt2n^{\frac32}}\) is convergent, thus \(\sum b_n\) is convergent
Finally, \(\sum a_n\) is convergent.