Proof of convergence of integral sequence using the definition of the limit

Question

"Let f(x) be a continuous, positive and decreasing function. Use the definition of the limit to prove that

$$\int_{1}^{∞}f(x)dx$$ converges ⇔ the sequence $$a_{n}=\int_{1}^{n}f(x)dx$$ converges."

I'm struggling with this proof. Can someone please guide me?

Answer 1

If this is your first proof-heavy class. One rule: always write down the definitions of the terms in your proof. It will point you in the direction of exactly what it is you need to be proving.

$\int_1^\infty f(x) dx = \lim_\limits {n\to\infty} \int_1^n f(x) dx$

The integral converges if the limit exists.

Suppose: $\lim_\limits {n\to\infty} \int_1^n f(x) dx = L$

$\forall \epsilon > 0, \exists N>0$ such that $n>N\implies |\int_1^n f(x) dx - L| < \epsilon$

If $n \in \mathbb N$: $\int_1^n f(x) dx = a_n$

$\forall \epsilon > 0, \exists N>0$ such that $n>N\implies |a_n - L| < \epsilon$

The sequence $\{a_n\}$ converges to $L$ if $\forall \epsilon>0,\exists N>0$ such than $n>N \implies |a_n-L|< \epsilon$

If the integral converges, the sequence converges.

To prove it the other way:

If sequence $\{a_n\}$ converges it is Cauchy.

$n,m>N \implies |a_n-a_m|< \epsilon\\ |\int_m^n f(x) dx| < \epsilon$

$f(x)$ positive is continuous and monotonically decreasing.

$0\le\int_m^r f(x)dx \le \int_m^n f(x)dx<\epsilon$ if $m<r<n$

note $m, n$ are integers while $r$ is real.

$m>N, r>m \implies |\int_m^r f(x)dx| < \epsilon$

Answer 2

I recommend starting by carefully writing out the definitions.

$$ \int_0^\infty f(t)dt := \lim_{x \to \infty} \int_0^x f(t)dt $$ What does it mean for this limit to converge to, say, $a \in \mathbb{R}$. It means for all $\epsilon>0$ there is a $c \in \mathbb{R}$ such that for all $x \geq c$, $|\int_0^x f(t)dt - a| < \epsilon \qquad (\ast)$

On the other hand, if $a_n := \int_0^n f(t)dt$, then the sequence $\{a_n\}_{n=1}^\infty$ converging to $a$ means for all $\epsilon>0$ there is a $k \in \mathbb{N}$ s.t for all $m \geq k$, $|a_m-a|<\epsilon$, i.e $|\int_0^m f(t)dt - a| < \epsilon$ The problem is asking you to investigate the difference between these two definitions-- the forward direction is (hopefully) rather straightforward, (fix $\epsilon>0$, if the first limit converges, you get your $c \in \mathbb{R}$ for which $(\ast)$ holds for [all] $x \geq c$, can you see a candidate for the aforementioned $k$?)

The reverse direction is a little more tricky and will involve invoking the hypotheses on $f$ (i.e continuity and monotonicity). As a hint, try considering a function for which the proposition doesn't hold and investigate what's going wrong. I recommend getting a piece of paper and drawing the following function in the xy plane, first draw the points $(0,0), (\frac{1}{4},1), (\frac{1}{2},0), (\frac{3}{4},-1), (1,0)$, connect each point to the following by a straight line, you should now have two triangles of equal area, one below the x-axis, one above-- repeat this pattern for all $x \geq 0$. Now notice $\int_0^1 f(t)dt=0$ (the positive and negative areas are equal), in fact $\int_0^n f(t)dt=0$ for all $n$, and hence the sequence $\{a_n\}$ clearly converges. But does $\lim_{x \to \infty} \int_0^x f(t)dt$ exist? Why not?

Good luck!