Theorem Suppose $$\begin{align}(a)&\sum_{n=0}^\infty a_n\quad\text{converges absolutely}\\(b)&\sum_{n=0}^\infty a_n=A\\(c)&\sum_{n=0}^\infty b_n=B\\(d)&c_n=\sum_{k=0}^n a_k b_{n-k}\quad(n=0,1,2,\dots)\end{align}$$ Then $$\sum_{n=0}^\infty c_n=AB$$

Proof. Put $$A_n=\sum_{k=0}^n a_k,\quad B_n=\sum_{k=0}^nb_k,\quad C_n=\sum_{k=0}^nc_k,\quad \beta_n=B_n-B.$$

Then after some manipulations we obtain $$C_n=A_nB+\underbrace {a_0\beta_n+a_1\beta_{n-1}+\dots+a_n\beta_0}_{\gamma_n}$$

We wish to show that $C_n\to AB$. Since $A_nB\to AB$, we only need to show that $\gamma_n\to0$ as $n\to \infty$. Then the proof proceeds as below. Put $$\alpha=\sum_{n=0}^\infty|a_n|.$$ Let $\epsilon>0$ be given. By (c), $\beta_n\to0$. Hence we can choose N such that $|\beta_n|\leq\epsilon$ for $n\ge N$, in which case

$$\begin{align}|\gamma_n|&\leq|\beta_0a_n+\dots+\beta_Na_{n-N}|+|\beta_{N+1}a_{n-N-1}+\dots+\beta_na_0| \\&\leq|\beta_0a_n+\dots+\beta_Na_{n-N}|+\epsilon\alpha\qquad(1)\end{align}$$

Keeping $N$ fixed, and letting $n\to\infty$, we get $$\limsup_{n\to \infty}|\gamma_n|\leq\epsilon\alpha$$ since $a_k\to0$ as $k\to\infty$. Since $\epsilon$ is arbitrary, we have proved the theorem.

Question: I can't understand the passage (1) and what follows. Could you please help me? Any hint or answer is welcome.

Thank you.


2 Answers 2


You reach $$ |\gamma_n|\leqslant |\beta_0a_n+\dots+\beta_na_{n-N}|+\epsilon\alpha$$

There is a typo, certainly, and it should read $$ |\gamma_n|\leqslant |\beta_0a_n+\dots+\color{red}{\beta_N}a_{n-N}|+\epsilon\alpha$$

Now, we have $N$ fixed summands. Taking the limit superior, since the right hand side converges, given for each $a_{n-k}\to 0$ as $n\to\infty$, you get $$\limsup\limits_{n\to\infty}|\gamma_n|\leqslant \varepsilon\alpha$$

Since $\varepsilon$ is arbitrary, you get $\limsup\limits_{n\to\infty}|\gamma_n|\leqslant 0$. Since $|\gamma_n|\geqslant 0$, we always have $\limsup\limits_{n\to\infty}|\gamma_n|\geqslant 0$, hence $\limsup\limits_{n\to\infty}|\gamma_n|= 0$. But $0\leqslant \liminf\limits_{n\to\infty}|\gamma_n|\leqslant \limsup\limits_{n\to\infty}|\gamma_n|= 0$, so $\lim\limits_{n\to\infty}|\gamma_n|=0$, so $\gamma_n\to 0$ as we wanted.

  • 1
    $\begingroup$ Thank you Pedro for your answer. How do I know that $$|\beta_{N+1}a_{n-N-1}+\dots+\beta_na_0|\leq\epsilon\alpha?$$ $\endgroup$
    – Charlie
    Feb 8, 2014 at 17:20
  • 4
    $\begingroup$ Note that it is $\leqslant \varepsilon|a_{n-N-1}+\cdots +a_0|$ for $\beta_k<\varepsilon$ if $k>N$. But $|a_0+\cdots+a_{n-N-1}|\leqslant |a_0|+\cdots+|a_{n-N-1}|\leqslant \sum_{n\geqslant 0}|a_n|=\alpha$. $\endgroup$
    – Pedro
    Feb 8, 2014 at 17:24
  • 3
    $\begingroup$ @Charlie Saying $x\leqslant \varepsilon$ for every $\varepsilon >0$ is equivalent to saying $x\leqslant 0$. $\endgroup$
    – Pedro
    Feb 8, 2014 at 17:47
  • $\begingroup$ Why does Rudin take the limsup rather than just the limit, since we know $a_k$ is a convergent sequence? $\endgroup$ Jun 3, 2022 at 4:27

Starting from the words "It is here that we use" on page 75, the proof basically states the following.

For converging series the terms converge to zero by Theorem 3.23 Cauchy criterion on page 60. In our case

\begin{align} n \to \infty \quad \text{ implies } \quad a_n, b_n \to 0 \end{align}

By the same token, as the partial sums $ \sum_{k=0}^{n}b_k = B_n$ converge to $B$ as $n \to \infty $, the "tails" between partial sums $B_n$ and $B$ (called $\beta_n$, i.e. $\beta_n=B_n-B$) also

\begin{align} n \to \infty \quad \text{ implies } \quad \beta_n \to 0 \end{align}

By following the steps on the proof, we arrived at the key inequality

\begin{align} \left| \gamma_n \right| \le &\underbrace{ \left| \beta_0 a_n + \dots + \beta_N a_{n-N} \right| }_\text{terms ending in $\beta_N$} + \underbrace{ \left| \beta_{N+1} \alpha_{n-N-1} + \cdots + \beta_n a_0 \right| }_\text{each $\beta$ having index over $N$, so each less than $\epsilon$ } = \nonumber \\ & \left| \beta_0 a_n + \dots + \beta_N a_{n-N} \right| + \underbrace{ \left| \beta_n a_0 \cdots \beta_{N+1} \alpha_{n-N-1} \right| }_\text{ flipped the order of terms} \nonumber \\ \le & \underbrace{ \left| \beta_0 a_n + \dots + \beta_N a_{n-N} \right| }_\text{Comment(**) below} + \underbrace{ \epsilon \cdot \alpha }_\text{ Comment (*) below } \nonumber \end{align}

Comment (*) :

  1. $\alpha$ contains infinite terms, whereas we had finitely many previously, $a_{n-N}$ terms in the right summand above (indexes run from 0 to $n-N-1$, no $n-N$ terms in total).

  2. all $ \beta $-terms are less than or equal $ \epsilon $.

  3. triangle inequality.

So in the last inequality we fixed the right part, namely $\epsilon \cdot \alpha $ It depends only on our arbitrary choice of $\epsilon$ (it thus can be as small as we wish), and inherent properties of the series $\sum a_n$, which converge absolutely to $\alpha$ that is also given.

Now, the left term (**), \begin{align} \left| \beta_0 a_n + \dots + \beta_N a_{n-N} \right| \label{key_term} \end{align}.

Here Rudin says "keep N fixed and push $n \to \infty $. This means the number of terms is fixed and remain the same (total N+1, as index with beta runs from zero to N), so the number of $\beta$'s that we add together remains the same despite we increase $n$! Only $a_n$ have index n in them, not betas!

Still, as we start to push n to infinity, indexes with a's all uniformly shift and increased by 1 and further to infinity. Their values during this process might be all reduced to zero. That is why the whole term goes to zero, as required.


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