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11.7 Series

Series

Theorem

If \(\sum_{n=1}^{\infty}\alpha^{n}\) is convergent, then \(a_{n}\xrightarrow{n\to\infty}0\)

Proof

By hypothesis we know that \(S_n=\sum_{n=1}^{\infty}\alpha^{n}\) is a convergent sequence, let \(l=\lim_{n\to\infty}S_{n}\)

But we know that \(S_n=S_{n-1}+a_n\), then \(a_n=S_n-S_{n-1}\) and \(\lim_{n\to\infty}S_{n}=l\Rightarrow\lim_{n\to\infty}S_{n-1}=l\)

Then \(\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}S_{n}-\lim_{n\to\infty}S_{n-1}=l-l=0\)

Corollary

If \(a_n\) is not convergent, then \(\sum_{n=1}^{\infty}\alpha^{n}\) is not convergent.

Remark

The converse is FALSE

For example, \(a_n=\frac1n\) and \(a_n\to 0\) is convergent, but \(\sum_{n=1}^{\infty}\frac{1}{n}=\infty\) is not convergent

Definition

We say that a series \(\sum_{n=1}^{\infty}\alpha^{n}\) is absolutely convergent if \(\sum_{n=1}^{\infty}|\alpha^{n}|\) is convergent

Example

If \(a_{n}=1,-\frac12,\frac13,-\frac14,\ldots\) which is \(a_{n}=\frac{\left(-1\right)^{n+1}}{n}\) it can be proved that \(\sum_{n=1}^{\infty}\alpha^{n}=\log\left(2\right)=\ln\left(2\right)\)

Thus \(\sum_{n=1}^{\infty}\alpha^{n}\) is convergent but not absolutely convergent since \(|a_n|=\frac1n\) and \(\sum_{n=1}^{\infty}\frac{1}{n}=\infty\)​

Theorem

If \(\sum_{n=1}^{\infty}\alpha^{n}\) is absolutely convergent, then \(\sum_{n=1}^{\infty}\alpha^{n}\) is convergent.

(The converse is false: The previous example shows a convergent series that is not absolutely convergent)

Example

\(a_{n}=\begin{cases}\frac{1}{2^{n}}~~~~\text{ if n is not prime}\\ -\frac{1}{2^{n}}~\text{ if n is prime}\end{cases}\) which is \(a_{n}=(\frac12,-\frac14,-\frac18,\frac{1}{16},-\frac{1}{32},\frac{1}{64},-\frac{1}{128},\ldots)\)

Is \(\sum_{n=1}^{\infty}a^{n}\) convergent?

Yes, \(|a_{n}|=\frac{1}{2^{n}}\) and \(\sum_{n=1}^{\infty}\left|a_{n}\right|=\sum_{n=1}^{\infty}\frac{1}{2^{n}}=1\Rightarrow\) is absolutely convergent

By theorem, \(\sum_{n=1}^{\infty}a_{n}\) is convergent.

Proof

Hypothesis: \(\sum_{n=1}^{\infty}\alpha^{n}\) is absolutely convergent, then \(\sum_{n=1}^{\infty}\left|\alpha^{n}\right|<\infty\) (convergent)

Then \(S_{n}=\sum_{k=1}^{n}\left|\alpha^{k}\right|\) is convergent. N.T.P. \(S_{n}^{\prime}=\sum_{k=1}^{n}\alpha^{k}\) is convergent

Then \(\forall\varepsilon>0,\exists n_0:n>n_0\Rightarrow\left|S_{n}-L\right|<\varepsilon\) and we want \(\forall\varepsilon>0,\exists n_0:n>n_0\Rightarrow\left|S_{n}^{\prime}-L\right|<\varepsilon\)

But we don't know the limit, thus we can use Cauchy Sequence.

​#Method tool: Cauchy Criterions to deal with unknown limit#​

Then we have \(\forall\varepsilon>0,\exists n_0:m,n>n_0\Rightarrow\left|S_{n}-S_{m}\right|<\varepsilon\).

Claim: \((S'_n)\) is also Cauchy Sequence

If \(n,m>n_0\), then \(\left|S^{\prime}_{n}-S^{\prime}_{m}\right|=\left|\sum_{k=1}^{n}\alpha^{k}-\sum_{k=1}^{m}\alpha^{k}\right|=\left|\sum_{k=m+1}^{n}\alpha^{k}\right|\leq\sum_{k=m+1}^{n}\left|\alpha^{k}\right|=\left|S_{n}-S_{m}\right|<\varepsilon\)​

Thus\((S'_n)\) is Cauchy Sequence, by A sequence is convergent if and only if it is a Cauchy sequence.

Thus \(\sum_{n=1}^{\infty}\alpha^{n}\) is convergent

Example 2

\(a_{n}=\frac12,\frac14,\frac18,\frac{1}{16},\frac{1}{32},\frac{1}{64},\frac{1}{128},\ldots\)image​

We cannot do that because after some periods, we never can reach the top from bottom since it is convergent

However, \(\frac1n\) can do that image​

Algebra of Series

Theorem

  1. If \(\sum_{n=1}^{\infty}a^{n}=A\) and \(\sum_{n=1}^{\infty}b^{n}=B\) are convergent, then \(\forall\alpha,\beta\in\mathbb{R},\sum_{n=1}^{\infty}\left(\alpha a_{n}+\beta b_{n}\right)\) is convergent and \(\sum_{n=1}^{\infty}\left(\alpha a_{n}+\beta b_{n}\right)=\alpha A+\beta B\)

    Proof

    Let \(A_{n}=\sum_{k=1}^{n}a^{k}\), \(B_{n}=\sum_{k=1}^{n}b^{k}\) and \(C_{n}=\sum_{k=1}^{n}\left(\alpha a_{k}+\beta b_{k}\right)\), hypothesis: \(\lim_{n\to\infty}A_{n}=A\) and \(\lim_{n\to\infty}B_{n}=B\)

    We want (Thesis): \(\lim_{n\to\infty}C_{n}=\alpha A+\beta B\)

    We know \(C_{n}=\sum_{k=1}^{n}\left(\alpha a_{k}+\beta b_{k}\right)=\sum_{k=1}^{n}\alpha a_{k}+\sum_{k=1}^{n}\beta b_{k}=\alpha A_{n}+\beta B_{n}\)

    From algebra of sequences we know: \(\lim_{n\to\infty}C_{n}=\alpha\lim_{n\to\infty}A_{n}+\beta\lim_{n\to\infty}B_{n}=\alpha A+\beta B\)

  2. If \(0\leq a_n\leq b_n\), then

    1. If \(\sum_{n=1}^{\infty}b^{n}<\infty\Rightarrow\sum_{n=1}^{\infty}a^{n}<\infty\)

      Let \(B_{n}=\sum_{k=1}^{n}b_{k}\). By hypothesis: \(\lim_{n\to\infty}B_{n}=B<\infty\)

      Let \(A_{n}=\sum_{k=1}^{n}a_{k}\). By hypothesis: \(a_k\leq b_k\), then \(A_n\leq B_n\)

      But \(A_n\) and \(B_n\) are monotonic increasing because \(a_n,b_n\geq 0\)

      Then \(B=sup(B_n)\), then \(B_{n}\leq B,\forall n\Rightarrow A_{n}\leq B_{n}\leq B\Rightarrow A_{n}\) is bounded above.

      Since \(A_n\) is also monotonic increasing, thus \(A_n\) is convergent \(\Rightarrow\) \(\sum_{n=1}^{\infty}a_{n}<\infty\)​ 2. If \(\sum_{n=1}^{\infty}a^{n}=\infty\Rightarrow\sum_{n=1}^{\infty}b^{n}=\infty\)​

      If \(\sum_{n=1}^{\infty}a^{n}=+\infty\), let \(A_{n}=\sum_{k=1}^{n}a_{k}\) and we have \(\lim_{n\to\infty}A_{n}=+\infty\)

      Thus \(\forall M>0,\exists n_0,\forall n>n_0,A_n>M\) (Hypothesis)

      We need to prove \(\forall M>0,\exists n_0,\forall n>n_0,B_{n}>M\) (Thesis)

      Since \(b_n\geq a_n\geq 0\), then \(B_n>A_n\). Then we use hypothesis here, \(B_n>A_n>M\)

      Therefore, \(\sum_{n=1}^{\infty}b^{n}=\infty\)​

Telescope sums

Examples

Is \(\sum_{n=1}^{\infty}\frac{1}{n^2}_{}<\infty\)?

Yes, because \(0\leq\frac{1}{n^2}<\frac{1}{n\left(n-1\right)},\forall n\geq2\)

If I prove that \(\sum_{n=1}^{\infty}\frac{1}{n\left(n-1\right)}_{}<\infty\), then by previous theorem \(\sum_{n=1}^{\infty}\frac{1}{n^2}_{}<\infty\)

Easily

However, we only can know \(\sum_{n=1}^{\infty}\frac{1}{n^2}_{}<\infty\), but we can not know what the exactly the limit is.

It's known that \(\sum_{n=1}^{\infty}\frac{1}{n^2}_{}=\frac{\pi^2}{6}\approx1.6\)

In general

\(\sum_{n=1}^{\infty}\frac{1}{n^2}_{}<\infty,\forall q>1.\sum_{n=1}^{\infty}\frac{1}{n^2}=\text{rational number}\times\pi^{q}\text{ for q even}\)

P-series

Consider \(\sum_{n=1}^{\infty}\frac{1}{n^{p}}\Rightarrow\begin{cases}\text{Convergent, if }p>1\\ \text{Divergent, if }p\leq1\end{cases}\)

Proof

When \(p>1\), we have already known \(\sum_{n=1}^{\infty}\frac{1}{n^2}_{}\) is convergent.

Then we use theorem: Since \(0<\frac{1}{n^{p}}\leq\frac{1}{n^2}_{}\left(p\geq2\right)\) and \(\sum_{n=1}^{\infty}\frac{1}{n^2}_{}<\infty\), then \(\sum_{n=1}^{\infty}\frac{1}{n^{p}}_{}<\infty\)

When \(p\leq 1\), we have already known \(\sum_{n=1}^{\infty}\frac{1}{n}_{}\) is divergent.

Then we use theorem: Since \(0<\frac{1}{n}\leq\frac{1}{n^{p}}_{}\left(p\leq1\right)\) and \(\sum_{n=1}^{\infty}\frac{1}{n^{}}=\infty\), then \(\sum_{n=1}^{\infty}\frac{1}{n^{p}}_{}=\infty\)

Definition

If \(b_{n}=a_{n}-a_{n+1}\), we have \(\sum_{k=1}^{n}b_{k}=a_1-a_{n+1}\)

Then \(\lim S_{n}=\lim_{n\to\infty}\left(a_1-a_{n+1}\right)=a_1-\lim_{n\to\infty}a_{n+1}=a_1\)​

Proposition

If \(b_n=a_n-a_{n+1}\) and \(a_n\to 0\), then \(\sum_{k=1}^{\infty}b_{k}\) is convergent to \(a_1\)

Example

  1. \(\sum_{n=1}^{\infty} \frac{\cos(n^2 + \frac{1}{n})}{n^3} = a_n < \infty ?\)

    Yes \(\quad 0 \leq |a_n| \leq \frac{1}{n^3} \Rightarrow \sum_{n=1}^{\infty} \frac{1}{n^3} < \infty\)

    Sandwich \(\Rightarrow \sum_{n=1}^{\infty} |a_n| < \infty \Rightarrow \sum_{n=1}^{\infty} a_n\) is absolutely convergent

    \(\Rightarrow \sum_{n=1}^{\infty} a_n < \infty\)

  2. \(\sum_{n=1}^{\infty} \frac{17^n + 3^n}{5^{2n}} < \infty ?\)

    Since \(\left(5^2\right)^n = 25^n\), then \(\sum_{n=1}^{\infty} \left(\frac{17}{25}\right)^n + \sum_{n=1}^{\infty} \left(\frac{3}{25}\right)^n < \infty\)

    \(\frac{17}{25} < 1, \quad \frac{3}{25} < 1\)

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