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Let $f,g:[a,\infty] \to \mathbb{R}$ s.t $\int_a^{\infty}f(t)dt$ converges, and $g$ is bounded and monotonic. Then $\int_a^{\infty}f(t)g(t)dt$ converges.

My idea: $g$ is monotonic thus integrable on $[a, x] \forall x$, and $f$ is also integrable, so $f(t)g(t)$ is integrable on $[a, x] \forall x$.

$g$ is monotnic and bounded thus $\lim_{x \to \infty}g(x) = L \in \mathbb{R}$, thus there exists $M>0$ s.t $\forall x>M, L-1 \leq g(x) \leq L+1$.

choosing some $b>M$ we get: $\int_b^{x}f(t)(L-1)dt \leq \int_b^{x}f(t)g(t)dt \leq \int_b^{x}f(t)(L+1)dt$

Taking $x \to \infty$, the left and right expressions converge because $\int_b^{\infty}f(t)dt$ converges, thus $\int_b^{\infty}f(t)g(t)dt$ converges.

Also, $\int_a^{b}f(t)g(t)dt$ obviously converges, so we get $\int_a^{\infty}f(t)g(t)dt$ converges.

Other proofs I saw usually used mean-value theorem so I wanna make sure I'm not missing something, thanks

paxtibimarce
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    This does not prove Abel' s test. You have the problem that $f$ may not be nonnegative and you can't immediately conclude that $(L-1)f(x) \leqslant f(x)g(x) \leqslant (L+1) f(x)$ for all $x$. Even if your final inequality were true, the best you could get after taking limits is $$(L-1)\int_b^\infty f(t) , dt \leqslant \liminf_{x \to \infty} \int_b^x f(t) g(t) , dt \leqslant \limsup_{x \to \infty} \int_b^x f(t) g(t) , dt \leqslant (L+1)\int_b^\infty f(t) , dt,$$

    since the far right-hand side and far left-hand sides are unequal.

     – RRL Sep 14 '20 at 18:28
  • The best way in my opinion is to show that the Cauchy criterion is satisfied as given in this answer. Proving convergence in a general setting without a good guess for the limit usually follows this path. – RRL Sep 14 '20 at 18:32

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Let $M = \displaystyle \lim_{x \rightarrow \infty} g(x)$, as you well said, real M exists since g is bounded and monotonic. Then we can bound your integral as follows:

$\displaystyle \int_{a}^{\infty} f(t)g(t) dt \leq M \displaystyle \int_{a}^{\infty} f(t) dt$

And as the right side integral converges by hyposthesis, you obtained a finite bound, this implies that the left side integral converges too.

Pablo Valentin Cortes Castillo
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