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 If it also converges after $f$ has been replaced by $\lvert f \rvert$, it is said to converge absolutely.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?
