Definition. We say that the series \(\sum_{n=1}^{\infty}a_n\) converges to \(A \in \mathbb{R}\) and write \(\sum_{n=1}^{\infty}a_n=A\) if the associated sequence of its partial sums \[s_n=\sum_{k=1}^{n}a_k=a_1+\ldots+a_n \ _{\overrightarrow{n \to \infty}} \ A.\] If \((s_n)_{n \in \mathbb{N}}\) diverges the series \(\sum_{n=1}^{\infty}a_n\) is said to diverge.
Remark.
Saying that the series \(\sum_{n=1}^{\infty}a_n\) converges we understand that \(\left|\sum_{k=1}^{\infty}a_k\right|<\infty\).
Saying that the series \(\sum_{n=1}^{\infty}a_n\) diverges we understand that \(\left|\sum_{k=1}^{\infty}a_k\right|=\infty\).
Exercise. If \(0 \leq x<1\), then \(\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}.\) If \(x \geq 1\), the series diverges.
Solution. If \(x < 1\), then \[s_n=\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}\] and the result follows if we let \(n \to \infty\).
For \(x \geq1\) note that \[\underbrace{1+1+\ldots+1}_{n} \leq s_n.\]
We have \(\lim_{n \to \infty}n=+\infty\), thus \(\lim_{n \to \infty}s_n=+\infty.\)
Exercise. \[\sum_{n=1}^{\infty}\frac{1}{k^2}<\infty.\]
Solution. Because the terms in the sum are all positive the sequence \[s_n=\sum_{k=1}^{n}\frac{1}{k^2} \quad \text{ is increasing.}\]
We now show that \((s_n)_{n \in \mathbb{N}}\) is bounded.
The (MCT) will prove that the series converges.
To prove boundedness of \((s_n)_{n \in \mathbb{N}}\) we note that \[\begin{aligned} s_n&=1+\frac{1}{2 \cdot 2}+\frac{1}{3 \cdot 3}+\frac{1}{4 \cdot 4}+\ldots+\frac{1}{n \cdot n} \\&<1+\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{(n-1)n} \\ &= 1+\left(1-{\color{red}\frac{1}{2}}\right)+\left({\color{red}\frac{1}{2}}-{\color{blue}\frac{1}{3}}\right)+\left({\color{blue}\frac{1}{3}}-{\color{purple}\frac{1}{4}}\right)+\ldots+\left(\frac{1}{n-1}-\frac{1}{n}\right) \\&=2-\frac{1}{n}<2. \end{aligned}\]
Thus the limit \(\lim_{n \to \infty}s_n\) exists.$$\tag*{$\blacksquare$}$$
One can also prove that \({\color{blue}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}}\). This is also Euler’s result.
Harmonic series. \[\sum_{n=1}^{\infty}\frac{1}{n}=\infty.\]
Solution. Note that \[\begin{aligned} 1+\frac{1}{2}+&{\color{red}\left(\frac{1}{3}+\frac{1}{4}\right)}+{\color{blue}\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)}+{\color{purple}\left(\frac{1}{9}+\ldots+\frac{1}{16}\right)}+{\color{brown}\left(\frac{1}{17}+\ldots\right.} \\& \geq 1+\frac{1}{2}+{\color{red}2\cdot \frac{1}{4}}+{\color{blue}4 \cdot \frac{1}{8}}+{\color{purple}8\cdot \frac{1}{16}}+{\color{brown}16 \cdot \frac{1}{32}}+\ldots \\& =1+\frac{1}{2}+{\color{red}\frac{1}{2}}+{\color{blue}\frac{1}{2}}+{\color{purple}\frac{1}{2}}+{\color{brown}\frac{1}{2}}+\ldots =1+\lim_{n \to \infty}\frac{n}{2}=\infty. \end{aligned}\] Thus \(s_n=\sum_{k=1}^{n}\frac{1}{k} \ _{\overrightarrow{n \to \infty}}\ \infty\). $$\tag*{$\blacksquare$}$$
Cauchy Condensation Test. Suppose that \((b_n)_{n \in \mathbb{N}}\) is decreasing and \(b_n \geq 0\) for all \(n \in \mathbb{N}\). Then the series \[{\color{red}\sum_{n=1}^{\infty}b_n<\infty} \quad \text{ converges }\] iff the series \[{\color{blue}\sum_{n=1}^{\infty}2^n b_{2^n}<\infty} \quad \text{ converges.}\]
Proof. Let \[s_n=b_1+b_2+\ldots+b_n,\] \[t_k=b_1+2b_2+\ldots+2^kb_{2^k}.\]
For \(n<2^k\) one has \[\begin{aligned} s_n &\leq b_1+\overbrace{b_2+b_3}^{2}+\ldots+\overbrace{b_{2^k}+\ldots+b_{2^{k+1}-1}}^{2^k}\\&\leq b_1+2b_2+\ldots+2^k b_{2^k}=t_k. \end{aligned}\]
(*). so that \(s_n \leq t_k\) for \(n<2^k\).
If \(n>2^k\) one has \[\begin{aligned} s_n &\geq b_1+b_2+(b_3+b_4)+\ldots+(b_{2^{k-1}+1}+\ldots+b_{2^k}) \\& \geq \frac{1}{2}b_1+b_2+2b_4+\ldots+2^{k-1}b_{2^k}=\frac{1}{2}t_k. \end{aligned}\]
(**). Thus \(2s_n \geq t_k\) for \(n>2^k\).
By (*) and (**) the sequences \((s_n)_{n \in \mathbb{N}}\) and \((t_k)_{k \in \mathbb{N}}\) are either both bounded or both unbounded. $$\tag*{$\blacksquare$}$$
Corollary. The series \[\sum_{n=1}^{\infty}\frac{1}{n^p} <\infty \quad \text{ iff }\quad p>1\]
Proof. The sequence \(b_n=\frac{1}{n^p}\) is decreasing and \(b_n \geq 0\) for all \(n \in \mathbb{N}\). By the Cauchy condensation test we obtain \[\sum_{n=1}^{\infty}\frac{1}{n^p}<\infty \quad \iff \quad \sum_{n=1}^{\infty}\frac{2^n}{2^{pn}}<\infty.\]
But the latter converges provided that
\[\sum_{n=1}^{\infty}\frac{2^n}{2^{pn}}=\sum_{n=1}^{\infty}2^{(1-p)n}=\frac{1}{1-\frac{1}{2^{p-1}}}<\infty \quad \iff \quad p>1.\tag*{$\blacksquare$}\]
Theorem. \[\sum_{n=0}^{\infty}\frac{1}{n!}=e.\]
Proof. Let \(s_n=\sum_{k=0}^{n}\frac{1}{k!}\). Then
\(s_n<s_{n+1}\) for all \(n \in \mathbb{N}\),
\(s_{n}=\sum_{k=0}^{n}\frac{1}{k!}=1+1+\sum_{k=2}^{n}\frac{1}{k!}<2+\sum_{k=2}\frac{1}{2^{k-1}}<3\).
Thus the limit \(\lim_{n \to \infty}s_n\) exists.
Let \(t_n=\left(1+\frac{1}{n}\right)^{n}\), then \(\lim_{n \to \infty}t_n=e\). By the binomial theorem
\[t_n=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}{n \choose k}\frac{1}{n^k}.\].
Then \[\begin{aligned} t_n&=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}{n \choose k}\frac{1}{n^k} \\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots (n-k+1)}{k!}\frac{1}{n^k} \\&=1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\ldots\\&+\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdot \ldots \cdot \left(1-\frac{n-1}{n}\right) =\sum_{k=0}^{n}\frac{1}{k!}=s_n. \end{aligned}\] Thus
\[e=\lim_{n \to \infty}t_n \leq \lim_{n \to \infty}s_n.\]
Next if \(n \geq m\) \[t_n \geq 1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\ldots+\frac{1}{m!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{m-1}{n}\right).\]
Let \(n \to \infty\) keeping \(m\) fixed, we get \[e=\lim_{n \to \infty}t_n \geq \sum_{k=0}^{m}\frac{1}{k!}.\]
Letting \(m \to \infty\) we see \(\lim_{m \to \infty}s_m \leq e\). \[\lim_{m \to \infty}s_m=\lim_{m \to \infty}\sum_{k=0}^{m}\frac{1}{k!}=e.\] This completes the proof of the theorem. $$\tag*{$\blacksquare$}$$
We have \(s_n=\sum_{k=0}^{n}\frac{1}{k!}<e\) for all \(n \in \mathbb{N}\). Indeed \[\begin{aligned} e-s_n&=\sum_{k=n+1}^{\infty}\frac{1}{k!}=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots \\&=\frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\frac{1}{(n+2)(n+3)}+\ldots\right) \\&<\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\ldots\right) \\&\leq \frac{1}{(n+1)!}\frac{1}{1-\frac{1}{n+1}}= \frac{1}{(n+1)!}\frac{n+1}{n}=\frac{1}{n! n}. \end{aligned}\] Hence we conclude
The error estimate (*). \[0<e-s_n<\frac{1}{n! n}.\]
Theorem. The Euler number \(e\) is irrational.
Proof. Suppose \(e\) is rational. Then \({\color{red}e=\frac{p}{q}}\) where \(p,q \in \mathbb{N}\). By (*) we have \[0<q!(e-s_q)<\frac{1}{q}.\] By our assumption \[{\color{red}q!e\in\mathbb N \quad \text{ is an integer}.}\]
Since \[q!s_{q}=q!\left(1+1+\frac{1}{2!}+\ldots+\frac{1}{q!}\right) \in \mathbb{N},\] we see \(q!(e-s_q) \in \mathbb{N}\), but if \(q > 1\) and this is impossible since \[0<q!(e-s_q)<1 / q<1.\] Hence \(e\) must be irrational. $$\tag*{$\blacksquare$}$$
Algebraic limit theorem for series. If \(\sum_{k=1}^{\infty}a_k=A\) and \(\sum_{k=1}^{\infty}b_k=B\) then \[\sum_{k=1}^{\infty}(\alpha a_k+\beta b_k)=\alpha A+\beta B.\]
Proof. Let \(A_n=\sum_{k=1}^na_k\) and \(B_n=\sum_{k=1}^nb_k\). We know that \[\lim_{n \to \infty}A_n=A, \quad \text{ and } \quad \lim_{n \to \infty}B_n=B,\]
so \[\begin{aligned} \lim_{n \to \infty}\sum_{k=1}^{n}(\alpha a_k+\beta b_k)&=\lim_{n \to \infty}\alpha \sum_{k=1}^{n}a_k+\beta\sum_{k=1}^{n}b_k \\ &=\alpha \lim_{n \to \infty}A_n+\beta \lim_{n \to \infty}B_n=\alpha A+\beta B.\qquad\end{aligned}\qquad\blacksquare\]
Theorem. The series \(\sum_{k=1}^{\infty}a_k\) converges iff for every \(\varepsilon>0\) there is \(N_{\varepsilon} \in \mathbb{N}\) such that whenever \(n>m \geq N_{\varepsilon}\) it follows \[\left|\sum_{k=m+1}^{n}a_k\right|<\varepsilon.\]
Proof. Let \(s_n=\sum_{k=1}^{n}a_k\) and we show that \((s_n)_{n\in\mathbb N}\) is a Cauchy sequence. Observe that whenever \(n>m \geq N_{\varepsilon}\) then \[|s_n-s_m|=\left|\sum_{k=m+1}^n a_k\right|<\varepsilon.\] We now apply the Cauchy Criterion for sequences and we are done.$$\tag*{$\blacksquare$}$$
Theorem. If the series \(\sum_{k=1}^{\infty}a_k\) converges then \(\lim_{n \to \infty}a_n=0\).
Proof. Let \(\varepsilon>0\) be given. Apply the previous theorem with \(m=n-1\), then \[|a_n|=|s_n-s_{n-1}|<\varepsilon\] whenever \(n>N_{\varepsilon}\), and we are done. $$\tag*{$\blacksquare$}$$
Remark. But \(\lim_{n \to \infty}a_n=0\) does not imply \(\left|\sum_{k=1}^{\infty}a_k\right|<\infty\).
Consider \(a_n=\frac{1}{n} \ _{\overrightarrow{n \to \infty}} \ 0\), but \(\sum_{n=1}^{\infty}\frac{1}{n}=\infty\).
Exercise. Determine if the series \[\sum_{n=1}^{\infty}(-1)^{n}\left(1-\frac{1}{n^3}\right)^{n^2}\] diverges or converges.
Solution. Since \((n^3)_{n \in \mathbb{N}}\) is a subsequence of \((n)_{n \in \mathbb{N}}\) we have \[\lim_{n \to \infty}\left(1-\frac{1}{n^3}\right)^{n^3}=e^{-1},\] hence \(\lim_{n \to \infty}\left(1-\frac{1}{n^3}\right)^{n^2}=1\), and the limit \(\lim_{n \to \infty}(-1)^{n}\left(1-\frac{1}{n^3}\right)^{n^2}\) does not exist, so the series diverges.$$\tag*{$\blacksquare$}$$
Comparison test. Assume that sequences \((a_k)_{k \in \mathbb{N}}\) and \((b_k)_{k \in \mathbb{N}}\) satisfy \[0 \leq a_k \leq b_k \quad \text{ for all }\quad k \in \mathbb{N}.\]
If \(\sum_{k=1}^{\infty}b_k\) converges, then \(\sum_{k=1}^{\infty}a_k\) converges.
If \(\sum_{k=1}^{\infty}a_k\) diverges then \(\sum_{k=1}^{\infty}b_k\) diverges.
Proof. Both statements follows from the Cauchy Criterion for series: \[\left|\sum_{k=m+1}^{n}a_k\right|\leq \left|\sum_{k=m+1}^{n}b_k\right|.\] This completes the proof. $$\tag*{$\blacksquare$}$$
Exercise. Determine if the series \[\sum_{n=1}^{\infty}\frac{1}{n^2+\sqrt{n}+15}\] diverges or converges.
Solution. For all \(n \in \mathbb{N}\) we have \[\frac{1}{n^2+\sqrt{n}+15} \leq \frac{1}{n^2},\quad \text{ thus }\] \[\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty,\] hence \[\ \qquad \qquad \sum_{n=1}^{\infty}\frac{1}{n^2+\sqrt{n}+15}<\infty.\qquad \tag*{$\blacksquare$}\]
Exercise. Determine if the series \[\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}+\sqrt{n}+1}\] diverges or converges.
Solution. For all \(n \in \mathbb{N}\) we have \[\frac{1}{\sqrt[3]{n}+\sqrt{n}+1} \geq \frac{1}{3\sqrt{n}},\quad \text{ thus }\] \[\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}=\infty,\] hence \[\ \qquad \qquad \sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}+\sqrt{n}+1}=\infty.\qquad \tag*{$\blacksquare$}\]
Theorem. A series of nonnegative terms \(a_k \geq 0\) converges iff its partial sums form a bounded sequence.
Proof. If \(\sum_{k=1}^{\infty}a_k<\infty\) one sees that \[s_N=\sum_{k=1}^{N}a_k \leq M=\sum_{k=1}^{\infty}a_k<\infty.\]
Conversely, we also know that \(s_N \leq s_{N+1} \leq M\) for all \(N \in \mathbb{N}\). Then the limit \[\lim_{N \to \infty}s_{N}\] exists by the (MCT).$$\tag*{$\blacksquare$}$$
Root test. Given \(\sum_{n=1}^{\infty}a_n\) set \[{\color{blue}\alpha=\limsup_{n \to \infty}\sqrt[n]{|a_n|}}.\]
If \(\alpha<1\), then \(\sum_{n=1}^{\infty}a_n\) converges.
If \(\alpha>1\), then \(\sum_{n=1}^{\infty}a_n\) diverges.
If \(\alpha=1\), no information.
Proof. If \(\alpha<1\) we can choose \(\beta\) so that \(\alpha<\beta<1\) and the integer \(N \in \mathbb{N}\) so that \[\sqrt[n]{|a_n|}<\beta \quad \text{ for all }\quad n \geq N,\] since \[\alpha=\inf_{k \geq 1}\sup_{n \geq k}\sqrt[n]{|a_n|}<\beta.\]
For \(n \geq N\) we have \(|a_n|<\beta^n\), but \(\beta<1\), thus \(\sum_{n=1}^{\infty}\beta^n\) converges and the comparision test implies that \(\sum_{n=1}^{\infty}a_n\) converges as well.
If \(\alpha>1\) then there is \((n_k)_{k \in \mathbb{N}}\) so that \[{|a_{n_k}|}^{1/n_k} \ _{\overrightarrow{k \to \infty}}\ \alpha.\]
Hence \(|a_n|>1\) holds for infinitely many values of \(n \in \mathbb{N}\), so that the condition \(a_n \ _{\overrightarrow{n \to \infty}}\ 0\) necessary for convergence \(\sum_{n=1}^{\infty}a_n\) does not hold.
To prove (c) note that \[\sum_{n=1}^{\infty}\frac{1}{n}=\infty \quad \text{ and } \quad \sqrt[n]{n} \ _{\overrightarrow{n \to \infty}}\ 1.\] \[\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty \quad \text{ and }\quad \sqrt[n]{n^2} \ _{\overrightarrow{n \to \infty}}\ 1.\] This completes the proof. $$\tag*{$\blacksquare$}$$
Example 1. \[\sum_{n=1}^{\infty}\left(\frac{e}{n}\right)^n<\infty,\] since \[\sqrt[n]{\frac{e^n}{n^n}}=\frac{e}{n}\ _{\overrightarrow{n \to \infty}}\ 0.\]
Example 2. \[\sum_{n=1}^{\infty}\frac{n^2}{2^n}<\infty,\] since \[\sqrt[n]{\frac{n^2}{2^n}}\ _{\overrightarrow{n \to \infty}}\ \frac{1}{2}.\]
Ratio test. The series \(\sum_{n=1}^{\infty}a_n\)
converges if \(\limsup_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|<1\),
diverges if \(\left|\frac{a_{n+1}}{a_n}\right| > 1\) for all \(n \geq n_0\) for some fixed \(n_0 \in \mathbb{N}\).
Proof. If (a) holds we can find \(\beta<1\) and \(n \in \mathbb{N}\) such that \[\left|\frac{a_{n+1}}{a_n}\right|<\beta \quad \text{ for all }\quad n \geq N.\]
In particular, for \(p \in \mathbb{N}\), one has \[\begin{aligned} |a_{n+p}|&=|a_{n+p-1}|\frac{|a_{n+p}|}{|a_{n+p-1}|}\\ &<\beta |a_{n+p-1}| \\&<\beta^2 |a_{n+p-2}|<\ldots< \\ &<\beta^p|a_n|. \end{aligned}\]
Thus \(|a_{N+p}|<\beta^p|a_N|\) and \[|a_n|<|a_N|\beta^{-N}\beta^n \quad \text{ for all } \quad n \geq N.\]
The claim follows from the comparison test since \(\sum_{n=1}^{\infty}\beta^n<\infty\) whenever \(\beta<1\).
If \(|a_{n+1}| \geq |a_n|\) for \(n \geq n_0\) then \(a_n \ _{\overrightarrow{n \to \infty}}\ 0\) does not hold. $$\tag*{$\blacksquare$}$$
Remark. As before \(\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=1\) is useless: \[{\color{red}\sum_{n=1}^{\infty}\frac{1}{n}=\infty\quad \text{ and } \quad \frac{a_{n+1}}{a_n}=\frac{n}{n+1} \ _{\overrightarrow{n \to \infty}}\ 1},\] \[{\color{blue}\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty\quad \text{ and }\quad \frac{a_{n+1}}{a_n}=\left(\frac{n}{n+1}\right)^2 \ _{\overrightarrow{n \to \infty}}\ 1.}\]
Example. \(\sum_{n=1}^{\infty}\frac{n!}{n^n}<\infty\), since \[\begin{aligned} \frac{a_{n+1}}{a_n}= \frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}=\frac{(n+1)n^n}{(n+1)^{n+1}}=\left(\frac{n}{n+1}\right)^n \ _{\overrightarrow{n \to \infty}}\ \frac{1}{e}<1. \end{aligned}\]