Solution. Let \(m>n\geq n_0\) we have \begin {align*} |S_m-S_n| &=\Big |\frac {(-1)^{n}}{n+1}+\cdots +\frac {(-1)^{m-1}}{m}\Big |\\ &=\Big |\frac {1}{n+1}-\frac {1}{(n+2)(n+3)}-\frac {1}{(n+4)(n+5)}-\cdots -\Big | \end {align*}

with negative last term \(\hspace {0.3cm}\dfrac {1}{m}<\dfrac {1}{n}<\dfrac {1}{n_0}\hspace {0.2cm},\hspace {0.3cm}\) choose \(n_0>\frac {1}{\varepsilon }\) then \(|S_m-S_n|<\varepsilon \) whenever \(n_0>\dfrac {1}{\varepsilon }\) so \((S_n)\) is Cauchy in \(\mathbb {R}\).

Proof. Suppose \((S_n)\) converges to both \(l\) and \(l'\). The strategy is to show \(d(l,l')\) is smaller than every positive number, which for a metric leaves only one possibility.

Let \(\varepsilon >0\). By convergence to \(l\) there is \(N_1\) with \(d(S_n,l)<\frac {\varepsilon }{2}\) for \(n\geq N_1\), and by convergence to \(l'\) there is \(N_2\) with \(d(S_n,l')<\frac {\varepsilon }{2}\) for \(n\geq N_2\). Take \(n\geq \max (N_1,N_2)\); then by the triangle inequality \[0\leq d(l,l')\leq d(l,S_n)+d(S_n,l')<\frac {\varepsilon }{2}+\frac {\varepsilon }{2} =\varepsilon .\] So \(0\leq d(l,l')<\varepsilon \) for every \(\varepsilon >0\), which forces \(d(l,l')=0\), and by the second metric axiom \(l=l'\).

Note where the axioms are used: the triangle inequality to compare the two limits through a common term, and \(d(x,y)=0\Rightarrow x=y\) to convert distance zero into equality. Without the second axiom the limit need not be unique, which is exactly what happens in a pseudometric space. □

Proof. Let \((S_n)\) be a convergent sequence in \((X,d)\). If the limit is not unique, then it may be \(l_1\) or \(l_2\). This means that \[d(S_n,l_1)<\frac {\varepsilon }{2}\hspace {0.5cm}\forall n\geq n_1\] \[d(S_n,l_2)<\frac {\varepsilon }{2}\hspace {0.5cm} \forall n\geq n_2\] Take \(n_0=\max \{n_1,n_2\}\), then we have \[0\leq d(l_1,l_2)\leq d(S_n,l_1)+d(S_n,l_2)<\frac {\varepsilon }{2}+\frac {\varepsilon }{2}=\varepsilon \hspace {0.3cm} \forall n\geq n_0\] \(d(l_1,l_2)\rightarrow 0\) as \(n\rightarrow \infty \). Thus \(l_1\) and \(l_2\) are but the same. □

Proof. Let \((S_n)\) be a convergent sequence in \((X,d)\). i.e \(\lim \limits _{n\rightarrow \infty }S_n=l\) where \(l\in (X,d)\). Then \[0\leq d(S_n,S_m)\leq d(S_n,l)+d(S_m,l)\rightarrow 0\hspace {0.3cm}\text {as}\hspace {0.3cm} n,m\rightarrow \infty .\] Hence \((S_n)\) is Cauchy.

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