= Solution
For a bounded real function on $[a,\infty)$, convergence of its <improper integral> means that its restriction to every $[a,b]$, $b>a$, is <Riemann integrable> and that the finite limit
$$
\boxed{\int_a^\infty f(x)\,dx
:=\lim_{b\to\infty}\int_a^b f(x)\,dx\in\mathbb R}
$$
exists. Boundedness by itself does not guarantee either local <Riemann integrability> or existence of this limit.
A <monotone function> on $[a,b]$ is bounded between its endpoint values. Divide the interval into $m$ equal pieces with endpoints $x_j=a+j(b-a)/m$. If $f$ is nondecreasing, its upper and lower <Darboux sums> differ by
$$
U(f,P)-L(f,P)
=\frac{b-a}{m}\sum_{j=1}^m\bigl(f(x_j)-f(x_{j-1})\bigr)
=\frac{b-a}{m}\bigl(f(b)-f(a)\bigr).
$$
For nonincreasing $f$, the same calculation with reversed endpoint values gives $(b-a)(f(a)-f(b))/m$. Thus in either case $U-L=(b-a)|f(b)-f(a)|/m\to0$. For every $\varepsilon>0$ one can choose $m$ to make this gap smaller than $\varepsilon$; the <Riemann integrability criterion> proves that $f$ is <Riemann integrable>.
Now let $f$ be decreasing on $[1,\infty)$ with limit zero. For each $x$, $f(x)\geq\lim_{y\to\infty}f(y)=0$. On $[n,n+1]$ its monotonicity gives
$$
f(n+1)\leq\int_n^{n+1}f(x)\,dx\leq f(n).
$$
Summing these bounds, with $S_N=\sum_{n=1}^Nf(n)$, yields
$$
\int_1^{N+1}f(x)\,dx\leq S_N
\leq f(1)+\int_1^Nf(x)\,dx.
$$
Both the <partial sums> $S_N$ and the truncated integrals are nondecreasing because $f\geq0$. The inequalities show that one family is bounded above if and only if the other is. The <bounded monotone sequence theorem> then gives convergence of the sums exactly when the integer-endpoint integrals converge. For real $b$ between $N$ and $N+1$, the integral to $b$ lies between the integrals to these integer endpoints; the <squeeze theorem> supplies the same limit. If either family is unbounded, both tend to $+\infty$. This proves the <integral test for convergence> in the required form.
For $\alpha<0$, use $f(x)=x^\alpha$, which is decreasing and tends to zero. Its truncated <improper integral> is
$$
\int_1^b x^\alpha\,dx=
\begin{cases}
\displaystyle\frac{b^{\alpha+1}-1}{\alpha+1},&\alpha\ne-1,\\[4pt]
\log b,&\alpha=-1.
\end{cases}
$$
This has a finite limit exactly when $\alpha<-1$. If $\alpha\geq0$, the terms $n^\alpha$ do not tend to zero, so the <term test for divergence> applies. Thus the full <p-series> criterion is
$$
\boxed{\sum_{n\geq1}n^\alpha\text{ converges }\iff\alpha<-1.}
$$
For every $\alpha\geq-1$ its nonnegative <partial sums> tend to $+\infty$.
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