Compare improper Riemann integrals to infinite series.

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Now we use integrals on unbounded domains to check whether series are
convergent or not.

**Theorem 1** (Integral Criterion for Series). *Let
\(f:[0,\infty)\to\mathbb{R}\) be
monotonically decreasing and non-negative. Then the series \[\sum_{k=0}^{\infty}f(k)\] is convergent if
and only if the integral \[\int_{0}^{\infty}f(x)dx\] converges. In
the case of convergence, the following estimate holds true: \[0\leq \sum_{k=0}^\infty f(k) -\int_0^\infty
f(x)~dx\leq f(0)\]*

*Proof:* Monotonicity of \(f\) implies that for \(k-1\leq x\leq k\) holds \(f(k)\leq f(x)\leq f(k-1)\). Monotonicity of
the integral therefore leads to the inequality \[f(k)=\int_{k-1}^kf(k)dx\leq
\int_{k-1}^kf(x)dx\leq \int_{k-1}^kf(k-1)dx=f(k-1).\] Therefore
\[\int_1^{n+2}f(x)dx\leq\sum_{k=1}^{n+1}
f(k)\leq \int_0^{n+1}f(x)dx\leq \sum_{k=0}^{n} f(k).\] Using this
inequality, we can directly conclude that, if one of the limits
(integral or sum) as \(n\to\infty\)
exists, then the other limit (sum or integral) also exists.

In case of convergence necessarily \((f(k))_{k\in\mathbb{N}_0}\) is a zero
sequence. Therefore \[0\leq \sum_{k=0}^n f(k)
-\int_0^{n+1} f(x)~dx\leq \sum_{k=0}^n f(k) - \sum_{k=1}^{n+1}f(k)
=f(0)-f(n+1)\] implies the inequality in Theorem 1 for \(n\rightarrow \infty\). \(\Box\)

**Example 3**.

Using the inproper integral of the function \(\int_0^\infty \frac{1}{x^\alpha} dx\), we
see that \[\sum_{k=1}^\infty\frac1{k^\alpha}\]
converges for \(\alpha>1\) and is
divergent for \(\alpha\leq1\).

For \(\alpha\in\mathbb{R}\)
consider \[\sum_{n=2}^\infty\frac1{n(\log(n))^\alpha}.\]
We have \[\begin{aligned}
\int_2^\infty\frac1{x(\log(x))^\alpha}~dx=&\lim_{n\to\infty}\begin{cases}\left.\frac{(\log(x))^{1-\alpha}}{1-\alpha}\right|_{x=2}^{x=n}&:\alpha\neq1,\\
\left.\log(\log(x))\right|_{x=2}^{x=n}&:\alpha=1\end{cases}\\
=&\begin{cases}-\frac{(\log(2))^{1-\alpha}}{1-\alpha}&:\alpha>1,\\
\infty&:\alpha\leq1.\end{cases}
\end{aligned}\] Therefore, the series \[\sum_{n=2}^\infty\frac1{n(\log(n))^\alpha}.\]
is convergent if and only if \(\alpha>1\).

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