Controlling Convergence

Let $f : [0, \infty) \to \mathbb R$ be continuous.

Problem 1. Suppose there exists $\eta > 0$ such that $\displaystyle \lim_{T \to \infty} T^{1+\eta}f(T) = 0$ . Evaluate

$\displaystyle \int_0^{\infty} tf'(t)\, \mathrm dt.$

(Click for Solution)

Solution. Integrating by parts,

$\displaystyle \int_0^T tf'(t)\, \mathrm dt = [tf(t)]_0^T - \int_0^T f(t)\, \mathrm dt = Tf(T) - \int_0^T f(t)\, \mathrm dt.$

We claim that each term on the right-hand side converges as $T \to \infty$ . For the first term, for $T \geq 1$ ,

$\begin{aligned} 0 \leq \lim_{T \to \infty} | Tf(T) | &\leq \lim_{T \to \infty} | Tf(T) | \cdot |T|^\eta \\ &\leq \lim_{T \to \infty} | T^{1+\eta} f(T) | = 0. \end{aligned}$

Therefore, $Tf(T) \to 0$ as $T \to \infty$ . For the integral, for $\epsilon := 1 > 0$ , there exists $K > 0$ such that

$\displaystyle t > K \quad \Rightarrow \quad |t^{1+\eta}f(t)| < \epsilon = 1.$

Setting $N := \max\{2, K\} \in \mathbb N$ ,

$\begin{aligned} \int_0^T |f(t)|\, \mathrm dt &= \int_0^N |f(t)|\, \mathrm dt + \int_N^T |f(t)|\, \mathrm dt \\ &=\int_0^N |f(t)|\, \mathrm dt + \int_N^T |f(t)| \cdot \frac{ |t|^{1+\eta} }{ |t|^{1+\eta} }\, \mathrm dt \\ &= \int_0^N |f(t)|\, \mathrm dt + \int_N^T \frac{ |t^{1+\eta} f(t)| }{ t^{1+\eta} } \, \mathrm dt \\ &\leq \int_0^N |f(t)|\, \mathrm dt + \int_N^T \frac{ 1 }{ t^{1+\eta} } \, \mathrm dt \\ &\leq \int_0^N |f(t)|\, \mathrm dt + \sum_{k=N}^{\infty} \frac{ 1 }{ k^{1+\eta} } \\ &\leq \int_0^N |f(t)|\, \mathrm dt + \sum_{k=1}^{\infty} \frac{ 1 }{ k^{1+\eta} }. \end{aligned}$

Taking $T \to \infty$ , since the series converges as a $p$ -series with $p = 1 + \eta > 0$ , since $f$ is continuous and thus integrable on $[0, N]$ , the integral on the left-hand side converges, and

$\displaystyle \left| \int_0^\infty f(t)\, \mathrm dt \right| \leq \int_0^\infty |f(t)|\, \mathrm dt < \infty.$

Therefore,

$\displaystyle \int_0^{\infty} tf'(t)\, \mathrm dt = -\int_0^{\infty} f(t)\, \mathrm dt.$

Problem 2. Suppose there exists $M_1, M_2 > 0$ such that $M_1 \leq f \leq M_2$ . Given $\eta \geq 0$ , determine the convergence of the series

$\displaystyle S_\infty := \sum_{k=1}^\infty \frac{f(k)}{k^{\eta}}.$

(Click for Solution)

Solution. Using the given estimate,

$\displaystyle 0 < \frac{M_1}{k^{\eta}} \leq \frac{f(k)}{k^{\eta}} \leq \frac{M_2}{k^{\eta}}.$

If $\eta > 1$ , then $\sum_{k=1}^\infty \frac{M_2}{k^{\eta}}$ converges. By the comparison test, $S_\infty$ converges. If $0 \leq \eta \leq 1$ , then $\sum_{k=1}^\infty \frac{M_1}{k^{\eta}}$ diverges. By the comparison test, $S_\infty$ diverges.

Problem 3. Suppose there exists $r > 0$ , $r \neq 1$ , such that

$\displaystyle \lim_{k \to \infty} \frac{g(k+1)}{g(k)} = r.$

Given $\eta > 0$ , determine the convergence of the series

$\displaystyle T_\infty := \sum_{k=1}^\infty \frac{g(k)}{k^{\eta}}.$

(Click for Solution)

Solution. Defining $a_k := g(k)/k^{\eta}$ , applying the criterion in the ratio test,

$\begin{aligned} \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| &= \lim_{k \to \infty} \left| \frac{g(k+1)}{(k+1)^{\eta}} \cdot \frac{k^{\eta}}{g(k)} \right| \\ &= \lim_{k \to \infty} \left( \left| \frac{g(k+1)}{g(k)} \right| \cdot \left( \frac{k}{k+1} \right)^{\eta} \right) \\ &= \left| \lim_{k \to \infty} \frac{g(k+1)}{g(k)} \right| \cdot \left( 1 - \lim_{k \to \infty} \frac{1}{k+1} \right)^{\eta} \\ &= |r| \cdot (1-0)^{\eta} = r. \end{aligned}$