Exercise 5.4.2

Question

Fix $x \in \mathbf{R}$. Argue that the series
$$
\sum_{n=0}^{\infty} \frac{1}{2^{n}} h\left(2^{n} x\right)
$$
converges and thus $g(x)$ is properly defined.

Ulisse Mini · Accepted Answer

$h_n(x) \leq 1/2^n$ and we know $\sum^\infty_{n=0} 1/2^n$ converges, so by the comparison test (Theorem 2.7.4) $g(x)$ is properly defined.

Exercise 5.4.2

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Exercise 5.4.2

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