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I am wrestling with the Cauchy integral criterion proof. In my math book (calculus) they prove this by the following:

The theorem:

Suppose that the continuous function $f(x)$ is positive and decreasing in the interval $ x\geq 1$

Then the series: $\sum_{k=1}^{\infty} f(x)$ is converging if, and only if the "generalized" integral $\int_{1}^{\infty} f(x)dx$ is converging.


$f(x)\geq 0 $ implies that $S_n=\sum_{k=1}^{n} f(k)$ is an increasing sequence, and has the limit $\lim_{n_\to \infty} S_n$ only if the sequence $S_n$ is upwards bounded.

Because of (1): $\int_{1}^{n} f(x)dx + f(n) \leq \sum_{k=1}^{n} f(k) \leq \int_{1}^{n} f(x)dx + f(1)$

(1) is developed by considering the sequence $f(2)*1+f(3)*1+...+f(n)*1=\sum_{k=2}^{n} f(k)$

where $\sum_{k=2}^{n} f(k) \leq \int_{1}^{n} f(x)dx$ and

$\int_{1}^{n} f(x)dx \leq f(1)*1+f(2)*1+...+f(n-1)*1 = \sum_{k=1}^{n-1} f(k)$

The inequalities above can be summarized to (1).

$S_n$ will be upward bounded exactly when $\int_{1}^{X} f(x)dx$ is upward bounded by the function $X$, i.e when the "generalized" integral $\int_{1}^{\infty} f(x)dx$ is converging.

My problem is that I dont really understand how they apply (1) and what they mean by the integral being bounded by the function X. So if anyone would be so kind to explain this in a bit more detail I would be more than happy!

Thank you!

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What book is that and on what page? I find the rightmost inequality in (1) a little strange... – DonAntonio Nov 17 '12 at 0:52
I can add some more detail to how they get to that inequality, it is from a Swedish book, so I guess it is not so well known :) – Lukas Arvidsson Nov 17 '12 at 0:54
up vote 1 down vote accepted

I assume, (1) means the hypothesis, that is, that $f$ is decreasing. So that, $$\int_1^2 f(x)dx \le \int_1^2 f(1) = f(1)$$ and this is used for all intervals $[k,k+1]$.

They mean "as the function of $X$", i.e. this map $X\mapsto \int_1^Xf(x)dx$ is bounded.

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Thank you for your answer! I might understand what you are getting at. Will try think think it through now. – Lukas Arvidsson Nov 17 '12 at 1:10
I think I got it now, thank you very much! – Lukas Arvidsson Nov 17 '12 at 1:26

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