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This is for improper integrals, but I believe the one for series would be similar.

Dirichlet test requires $f(x)$ to be monotone, $\lim_{x \to \infty}f(x) = 0$ and $\int_{a}^{b}g(x)$ to be bounded by constant $B$. Then $\int_{a}^{+\infty}f(x)g(x)dx$ converges.

Meanwhile, Abel test requires $f(x)$ to be monotone and bounded while $\int_{a}^{\infty}g(x)dx$ converges. Then $\int_{a}^{+\infty}f(x)g(x)dx$ converges.

I see how the two are very similar, except with the tighter/looser conditions on $f$ and $g$.

How do we derive the Abel test from the Dirichlet test?

RRL
  • 90,707
sedrick
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1 Answers1

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Both theorems are proved by applying the second mean value theorem for integrals (given that $f$ is monotone). For $c_2 > c_1 > a$, there exists $\xi \in (c_1,c_2)$ such that

$$\tag{*}\left|\int_{c_1}^{c_2}f(x) g(x) \, dx\right| = \left|f(c_1)\int_{c_1}^{\xi}g(x) \, dx + f(c_2)\int_{\xi}^{c_2}g(x) \, dx \right| \\ \leqslant |f(c_1)|\left|\int_{c_1}^{\xi}g(x) \, dx\right| + |f(c_2)|\left|\int_{\xi}^{c_2}g(x) \, dx\right|.$$

One theorem is not a corollary of the other. The hypotheses, while sharing some common characteristics, are independent and under each set of hypotheses the RHS of (*) is arbitrarily small when $c_1$ is sufficiently large, and the improper integral converges by the Cauchy criterion. With Dirichlet, the integrals on the RHS are uniformly bounded for all $a < c_1 < \xi < c_2$ and $f(c_1),f(c_2) \to 0$. With Abel, $f(c_1),f(c_2)$ are bounded for all $a < c_1 < c_2$ and the integrals are arbitrarily small when $c_1$ is sufficiently large since $\int_1^\infty g(x) \, dx$ converges.

Under the hypotheses for the Dirichlet test, we have $\int_a^b g(x) \,dx$ bounded for all $b > a$, but the improper integral $\int_a^\infty g(x) \, dx$ may not converge. We need $f$ not only to be monotone and bounded so that $\lim_{x \to \infty}f(x) = L$ exists, but also that $L =0$ as a sufficient condition.

RRL
  • 90,707