equivalent definitions of convergence of improper integrals Show $$\int_{a}^{\infty} f$$ if and only if $$\forall \epsilon>0,  \exists M>a$$ such that whenever $d>c\geq M$, it follows that $\lvert\int_{c}^{d}f \rvert<\epsilon$
prove this for the forwards direction ie. assume $\int_{a}^{\infty} f$ convergent and prove $\lvert\int_{c}^{d}f\rvert<\epsilon$.
(only forwards direction proof required,  I can prove backwards direction)

The question provides the hint to define $$a_n:=\left|\int_{a}^{a+n}f\right|$$
My attempt (forwards direction):
take $\int_{a}^{\infty}f$ as convergent, setting $a_n$ as above and take $m>n>M$, then $\lim_{n\to\infty}\int_{a}^{a+n} f$ is defined, $a_n$ is therefore convergent giving Cauchy convergence, hence, $\lvert a_n-a_m\rvert<\epsilon$.
$$\left| a_n-a_m\right|=\left|\int_{a}^{a+n}f-\int_{a}^{a+m}f\right|=\left|\int_{a+m}^{a+n}f\right|<\epsilon$$
But we have to prove $\left|\int_{c}^{d}f\right|<\epsilon$.
So I tried setting $a+n\leq c<d\leq a+m$ and breaking $\left|\int_{a+m}^{a+n}f\right|$ down using inequalities with no success. I am not sure how to continue.
Any help would be greatly appreciated. Thank you in advance.
 A: By hypothesis, for any $\epsilon > 0$ there exists $M'$ such that for all $d > c > M'$, we have
$$\left|\int_c^d f(x) \, dx \right| < \frac{\epsilon}{2}$$
Consider the sequence $\displaystyle A_n = \int_a^{a+n}f(x) \, dx$.  For all $m> n > N_1(\epsilon) = \lceil M'-a \rceil$, we have $a+n > M'$, and
$$\tag{1}|A_m -A_n | = \left|\int_{a+n}^{a+m}f(x)\, dx \right| < \frac{\epsilon}{2} < \epsilon$$
Thus, $(A_n)$ is a Cauchy sequence in $\mathbb{R}$ and there exists $I\in \mathbb{R} $ such that $A_n \to I$ as $n \to \infty$.
It remains to show that $\displaystyle I = \lim_{c \to \infty} \int_a^c f(x) \, dx = \int_a^\infty f(x) \, dx$.
Since $A_n \to I$ there exists $
N_2(\epsilon) \in \mathbb{N}$ such that for all $n > N_2(\epsilon)$,
$$\tag{2}\left|A_n - I \right| = \left|\int_a^{a+n} f(x) \, dx- I\right| < \frac{\epsilon}{2}$$
Fix $K = \max(N_1(\epsilon), N_2(\epsilon))$. For all $c > a+K$, we have
$$\left|\int_a^c f(x) \, dx - I \right| \leqslant \left|\int_a^{a+K} f(x) \, dx - I\right| +  \left|\int_a^{c} f(x) \, dx - \int_a^{a+K} f(x) \, dx \right| \\ =  \underbrace{\left|\int_a^{a+K} f(x) \, dx - I\right|}_{< \frac{\epsilon}{2}} +  \underbrace{\left|\int_{a+K}^{c} f(x) \, dx\right|}_{< \frac{\epsilon}{2}}< \epsilon $$
Note that the first term on the RHS is smaller than $\epsilon/2$ by (2) since $K > N_2(\epsilon)$.  The second term on the RHS is smaller than  $\epsilon/2$ by (1) since $c > a+K > a+ N_1(\epsilon)\geqslant M'$.
This proves the convergence of
$$\lim_{c \to \infty}\int_a^c f(x) \, dx = I$$
