While studying series I came on the following proof. But I don't really understand it.

The theorem is the following:

Theorem: Suppose that f: $[a,+ \infty) \rightarrow R_{+}^{}$ is a continuous and non-increasing function. Suppose that $p \in N$ is greater or equal to a. Then converges the integral \begin{align} \int_{a}^{+\infty} f(t)dt \end{align} if and only if the serie

\begin{align} \sum\limits_{n=p}^{\infty}f(n) \end{align} Converges.

The proof starts as following:

For $p \leq n \leq t \leq n+1$ is $f(p) \geq f(n) \geq f(t) \geq f(n+1)$*

And so

$ f(n+1) =_{}^{**} \int_{n}^{n+1}f(n+1)dt \leq \int_{n}^{n+1}f(t)dt \leq \int_{n}^{n+1}f(n)dt=f(n) $

Suppose now that the integral converges. Then the partial sum is:

$s_{n}^{} = f(p)+f(p+1)+...+f(n) \leq f(p) + \int_{p}^{p+1}f(t)dt_{}^{***} +...+ \int_{n}^{n-1}f(t)dt = f(p) + \int_{p}^{n}f(t)dt \leq f(p)+ \int_{p}^{+\infty}f(t)dt $

And so we see that the serie of the partial sum is bounded and the serie converges.

Suppose now that the $ \sum\limits_{n=p}^{\infty}f(n) $ converges with sum s. Take x $ \geq $ p and n $ \in N > x.$ Then

$ \int_{a}^{x}f(t)dt \leq \int_{a}^{n}f(t)dt = \int_{a}^{p}f(t)dt + \int_{p}^{p+1}f(t)dt+...+\int_{n-1}^{n}f(t)dt \leq \int_{a}^{p} +f(p)+...+f(n-1) \leq \int_{a}^{p}f(t)dt +s_{}^{****} $

So we proved that $ \int_{a}^{x}f(t)dt $ is an increasing and bounded function of x. So the integral exists.

I put stars where I don't really understood the step so:

*: I see why $ f(n) \geq f(n+1) $ ( f is non-decreasing but what about p and t?

**: Why is $ f(n+1) $ equal to $\int_{n}^{n+1}f(n+1)dt$?

***: Why is $ f(p+1) $ equal to $ \int_{p}^{p+1}f(t)dt $?

****: Isn't s equal to f(p)+f(p+1)+...+f(n-1)+f(n)?

Thanks in advance!


3 Answers 3


There are multiple issues so that it is too long to be typed in the comment.

For *, the definition of a non-increasing is:

$$a \leq b \implies f(a) \geq f(b)$$

So that if you apply this definition successively to the chain of inequalities $p\leq n \leq t \leq n+1$ (between $p, n$ then $n, t$ and so on), you get a reversed chain of inequality.

For **, remember that $f(n+1)$ is just a constant and the dummy variable is $t$. $\int_n^{n+1} f(n+1)\, dt = f(n+1) \int_{n}^{n+1}dt = f(n+1)$

For ***, the author didn't mean $f(p+1) = \int_p^{p+1} f(t)dt$, he actually meant $f(p+1) \leq \int_p^{p+1} f(t) dt$, $f(p+2) \leq \int_{p+1}^{p+2} f(t) dt $ and $f(n) \leq \int_{n-1}^{n} f(t) dt$ so adding these subinequalities up gives you the inequality in the proof. But do you know why each subinequality holds?

For ****, yes. But since $f(n) \geq 0$, $s = f(p) + \cdots + f(n-1) + f(n) \geq f(p) + \cdots + f(n-1) + f(n)$.

A general comment: I think being cautious is the first step in doing analysis. So you actually have a good start. Now you have identified the parts where you don't quite understand, or "gaps". Next you shall pick up a pencil and try filling in these gaps. Preferably, try "re-proof" in your own words with the previous theorems. Also, advanced books tend to omit more details and leave more chances for you to practice. So try to develop such a habit.

  • 1
    $\begingroup$ ***: The subinequalities holds because $ f(p+1)= \int_{p}^{p+1}f (p+1)dt \leq \int_{p}^{p+1}f (t)dt $ $\endgroup$ Jun 7, 2017 at 12:48

(*) It is basically a rewriting of the statement $f(x)\ge f(y)\iff x\le y$ for strictly increasing functions. We then chain multiple of these inequalities together:

$f(t)\ge f(n+1)\iff t\le n+1,\quad f(n)\ge f(t)\iff n\le t,\quad\dots$

(**) Notice that the integral is with respect to $t$, and $f(n+1)$ is constant for all $t$, so,

$$\int_n^{n+1}f(n+1)~\mathrm dt=f(n+1)\int_n^{n+1}1~\mathrm dt=f(n+1)$$

(***) It isn't the case that $f(p+1)=\displaystyle\int_p^{p+1}f(t)~\mathrm dt$, but rather that $f(p+1)\le\displaystyle\int_p^{p+1}f(t)~\mathrm dt$.

$$\int_p^{p+1}f(t)~\mathrm dt\ge\int_p^{p+1}f(p+1)~\mathrm dt\iff p+1\ge t$$

As per the bounds, it is true that $p+1\ge t$. Using the previous (**), we note that

$$\int_p^{p+1}f(p+1)~\mathrm dt=f(p+1)\int_p^{p+1}1~\mathrm dt=f(p+1)$$

As is expected.

(****) Yes, indeed, that is how $s$ is defined. As we can see,

$$\int_{a}^{p}f(t)~\mathrm dt +\underbrace{f(p)+\dots+f(n-1)}_{s-f(n)\le s} \leq \int_{a}^{p}f(t)~\mathrm dt +s_{}^{****}$$

And I hope that clears everything up for you.



You just need to know that


If $g $ is a non-increasing function at $[a,b] $ then if $$a\le c\le d\le b $$ then $$g (a)\ge g (c) \ge g (d)\ge g (b) $$


$$\int_a^b\lambda dt=\lambda\int_a^bdt =\lambda (b-a)$$ or

$$\int_p^{p+1}f (p+1)dt=$$ $$f (p+1)\Big [(p+1)-p\Big]=f (p+1) $$


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