# Prob. 8, Chap. 6, in Baby Rudin: The Integral Test for Convergence of Series

Here is Prob. 8, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f \in \mathscr{R}$ on $[a, b]$ for every $b > a$ where $a$ is fixed. Define $$\int_a^\infty f(x) \ \mathrm{d} x = \lim_{b \to \infty} \int_a^b f(x) \ \mathrm{d} x$$ if this limit exists (and is finite). In that case, we say that the integral on the left converges. If it also converges after $f$ has been replaced by $\lvert f \rvert$, it is said to converge absolutely.

Assume that $f(x) \geq 0$ and that $f$ decreases monotonically on $[1, \infty)$. Prove that $$\int_a^\infty f(x) \ \mathrm{d} x$$ converges if and only if $$\sum_{n=1}^\infty f(n)$$ converges. (This is the so-called "integral test" for convergence of series. )

My Attempt:

As $f(x) \geq 0$ on $[1, \infty)$, so for each $n\in \mathbb{N}$ we see that $$\sum_{k=1}^{n+1} f(k) = \sum_{k=1}^n f(k) \ + \ f(n+1) \geq \sum_{k=1}^n f(k). \tag{0}$$ Thus the sequence $\left( \sum_{k=1}^n f(k) \right)_{n \in \mathbb{N} }$ is a monotonically increasing sequence of (non-negative) real numbers.

Suppose $b$ and $c$ are any two real numbers such that $1 < b < c$. Then we have \begin{align} \int_1^c f(x) \ \mathrm{d} x &= \int_1^b f(x) \ \mathrm{d} x + \int_b^c f(x) \ \mathrm{d} x \qquad \mbox{ [ by Theorem 6.12 (c) in Baby Rudin ] } \\ &\geq \int_1^b f(x) \ \mathrm{d} x + 0 \\ & \ \ \ \qquad \mbox{ [ by Theorem 6.12 (b) in Baby Rudin since f(x) \geq 0 on [b, c] ] } \\ &= \int_1^b f(x) \ \mathrm{d} x. \tag{1} \end{align} which shows that the function $g$ defined on $(1, \infty)$ by $$g(b) = \int_1^b f(x) \ \mathrm{d} x \tag{A}$$ is a monotonically increasing function.

Now suppose that $b$ is a real number such that $b > 2$. Let $n$ be the natural number such that $n \leq b < n+1$. Then we see that \begin{align} \int_1^b f(x) \ \mathrm{d} x &\leq \int_1^{n+1} f(x) \ \mathrm{d} x \qquad \mbox{ [ using (1) above; note that n \leq b < n+1  ] } \\ &= \sum_{k=1}^n \int_k^{k+1} f(x) \ \mathrm{d} x \qquad \mbox{ [ by an extension of Theorem 6.12 (c) in Rudin ] } \\ &\leq \sum_{k=1}^n \int_k^{k+1} f(k) \ \mathrm{d} x \\ &\ \ \ \qquad \mbox{ [ using Theorem 6.12 in Baby Rudin and the monotonicity of f ] } \\ &= \sum_{k=1}^n f(k). \tag{2} \end{align} And, also \begin{align} \int_1^b f(x) \ \mathrm{d} x &\geq \int_1^n f(x) \ \mathrm{d} x \qquad \mbox{ [ using (1) above; note that n \leq b < n+1  ] } \\ &= \sum_{k=1}^{n-1} \int_k^{k+1} f(x) \ \mathrm{d} x \qquad \mbox{ [ by an extension of Theorem 6.12 (c) in Rudin ] } \\ &\geq \sum_{k=1}^{n-1} \int_k^{k+1} f(k+1) \ \mathrm{d} x \\ &\ \ \ \qquad \mbox{ [ using Theorem 6.12 in Baby Rudin and the monotonicity of f ] } \\ &= \sum_{k=1}^{n-1} f(k+1) \\ &= \sum_{k=2}^n f(k). \tag{3} \end{align}

Thus from (2) and (3) we can conclude that for every real number $b > 2$, we have $$\sum_{k=2}^n f(k) \leq \int_1^b f(x) \ \mathrm{d} x \leq \sum_{k=1}^n f(k), \tag{4}$$ where $n$ is the natural number such that $n \leq b < n+1$.

Now suppose that $\int_1^\infty f(x) \ \mathrm{d} x$ converges. This means that $\lim_{b \to \infty} \int_1^b f(x) \ \mathrm{d} x$ exists in $\mathbb{R}$, and in the light of (1) we can also write
$$\int_1^\infty f(x) \ \mathrm{d} x = \lim_{b \to \infty} \int_1^b f(x) \ \mathrm{d} x = \sup \left\{ \ \int_1^b f(x) \ \mathrm{d} x \ \colon \ b \in \mathbb{R}, \ b > 1 \ \right\}. \tag{5}$$

Then for any natural number $n \geq 2$, if we take $b \in (n, n+1)$, then from (4) and (5) we can conclude that $$\sum_{k=2}^n f(k) \leq \int_1^\infty f(x) \ \mathrm{d} x,$$ which implies that $$\sum_{k=1}^n f(k) \leq f(1) \ + \ \int_1^\infty f(x) \ \mathrm{d} x.$$ Thus the sequence $\left( \sum_{k=1}^n f(k) \right)_{n \in \mathbb{N}}$ of (non-negative) real numbers is bounded above, and (0) shows that this sequence is also monotonically increasing; therefore this sequence is convergent (in $\mathbb{R}$). That is, the series $\sum f(n)$ is convergent.

Conversely, suppose that the series $\sum f(n)$ is convergent. Then the sequence $\left( \sum_{k=1}^n f(k) \right)_{n \in \mathbb{N}}$ converges in $\mathbb{R}$; but by (0) above this is a monotonically increasing sequence of real numbers; so it is a bounded above sequence and $$\sum_{n=1}^\infty f(n) = \lim_{n \to \infty} \sum_{k=1}^n f(k) = \sup \left\{ \ \sum_{k=1}^n f(k) \ \colon \ n \in \mathbb{N} \ \right\}. \tag{6}$$

Now let $b$ be any real number such that $b > 2$, and let $n$ be the natural number such that $n \leq b < n+1$. Then from (4) and (6) we can conclude that $$\int_1^b f(x) \ \mathrm{d} x \leq \sum_{k=1}^n f(k) \leq \sum_{n=1}^\infty f(n).$$

Thus the function $g$ given by (A) above is a monotonically increasing function on $(2, \infty)$ which is also bounded above. So $\lim_{b \to \infty} g(b)$ exists in $\mathbb{R}$; that is $\lim_{b \to \infty} \int_1^b f(x) \ \mathrm{d} x$ exists in $\mathbb{R}$, which is the same as saying that $\int_1^\infty f(x) \ \mathrm{d} x$ converges.

Is my proof correct? If so, then have I managed to present it in enough detail and rigorous for it to be understood by someone who is not very sharp (like myself) and is also taking their very first course in analysis?

Or, have I committed some blunders?

• Soon-to-be grad student here with decent experience in analysis, I like it. If there is a mistake, I can't find it and the presentation is very clear and easy to follow! – Laplacinator Jul 31 '17 at 13:12
• @Chris thanks. Where have you had your very first course in analysis? And what textbook, if any, were you taught from? – Saaqib Mahmood Jul 31 '17 at 13:23
• My first course in analysis was at Umeå University here in Sweden, and we used Baby Rudin! – Laplacinator Jul 31 '17 at 14:32
• @Chris great! Lucky you! Was that course in English? Have you got any materials from that course? And if so, can you share it with me plesse? – Saaqib Mahmood Jul 31 '17 at 16:57
• Hi again Saaqib. I agree with Chris here, that this is solid. You wrote more words about this exercise than I did, but I used exactly the same ideas. I have a propensity to hand-wave probably more than I should, but it sure seems to me like this is good. I think another way to do it is with the Cauchy criteria for each. You can bound each in terms of some partial sum for the other. – Alfred Yerger Aug 1 '17 at 3:59