Prove the following standard facts about infinite series:
Show that if \((s_n)\) is a bounded sequence of real numbers and \(s_n \leq s_{n+1}\) for each \(n\), then \((s_n)\) converges.
Let \((a_n)\) be a sequence of real numbers; define
\[ s_n = \sum_{i=1}^{n} a_i. \]
If \(s_n \to s\), we say that the infinite series
\[ \sum_{i=1}^{\infty} a_i \]
converges to \(s\) also. Show that if \(\sum a_i\) converges to \(s\) and \(\sum b_i\) converges to \(t\), then \(\sum (ca_i + b_i)\) converges to \(cs + t\).
Prove the comparison test for infinite series: if \(\abs{a_i} \leq b_i\) for each \(i\), and if the series \(\sum b_i\) converges, then the series \(\sum a_i\) converges. [Hint: Show that the series \(\sum \abs{a_i}\) and \(\sum c_i\) converge, where \(c_i = \abs{a_i} + a_i\).]
Given a sequence of functions \(f_n: X \to \mathbb{R}\), let
\[ s_n(x) = \sum_{i=1}^{n} f_i(x). \]
Prove the Weierstrass \(M\)-test for uniform convergence: if \(\abs{f_i(x)} \leq M_i\) for all \(x \in X\) and all \(i\), and if the series \(\sum M_i\) converges, then the sequence \((s_n)\) converges uniformly to a function \(s\). [Hint: Let \(r_n = \sum_{i=n+1}^{\infty} M_i\). Show that if \(k > n\), then \(\abs{s_k(x) - s_n(x)} \leq r_n\); conclude that \(\abs{s(x) - s_n(x)} \leq r_n\).]
Solution
1. (a) A bounded nondecreasing sequence \((s_n)\) converges to \(s=\sup_ns_n\).
The supremum exists by the least upper bound property of \(\mathbb R\). Given \(\varepsilon>0\), \(s-\varepsilon\) is not an upper bound, so \(s_N>s-\varepsilon\) for some \(N\). For \(n\ge N\), \(s-\varepsilon<s_N\le s_n\le s\), so \(\abs{s_n-s}<\varepsilon\).
2. (b) If \(\sum a_i=s\) and \(\sum b_i=t\), then \(\sum(ca_i+b_i)=cs+t\).
With \(s_n=\sum_{i\le n}a_i\) and \(t_n=\sum_{i\le n}b_i\), the \(n\)-th partial sum of \(\sum(ca_i+b_i)\) is \(cs_n+t_n\), which converges to \(cs+t\) by the limit laws for sequences.
3. (c) If \(\abs{a_i}\le b_i\) and \(\sum b_i\) converges, then \(\sum a_i\) converges.
1.Since \(a_i=c_i-\abs{a_i}\), step 2 with \(c=-1\) shows that \(\sum a_i\) converges.
4. (d) If \(\abs{f_i(x)}\le M_i\) for all \(x\) and \(i\) and \(\sum M_i\) converges, then \((s_n)\) converges uniformly on \(X\).
For each \(x\), step 3 shows that \(\sum_if_i(x)\) converges; call its sum \(s(x)\). Let \(r_n=\sum_{i>n}M_i\), so \(r_n\to0\). For \(k>n\) and \(x\in X\), \[ \abs{s_k(x)-s_n(x)}\le\sum_{i=n+1}^k\abs{f_i(x)}\le\sum_{i=n+1}^kM_i\le r_n, \] and letting \(k\to\infty\) gives \(\abs{s(x)-s_n(x)}\le r_n\). Given \(\varepsilon>0\), choose \(N\) with \(r_N<\varepsilon\); then \(\abs{s(x)-s_n(x)}\le r_n\le r_N<\varepsilon\) for all \(n\ge N\) and all \(x\in X\).
5. Q.E.D.
Steps 1 through 4 prove (a) through (d).