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Suppose $\sum_{n=1}^{\infty}f(n)$ converges.
$0 \le f(x)$ on [1,$\infty$] and $\int_{1}^{N}f(x)dx < \sum_{n=1}^{\infty}f(n) < \infty$.

Define a sequence $a_N=\int_{1}^{N}f(x)dx$ then $a_N$ is bounded above and increasing,
therefore converges.

In here, I want to conclude $\int_{1}^{\infty}f(x)dx$ also converges.
It is possible that just applying limit and saying it is true for $\infty$?
Or is there any other theorem that I can use?

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Why would $(a_N)$ be bounded ? – beauby Nov 26 '12 at 16:18
From the first line, $a_N=\int_{1}^{N} f(x)dx < \infty$. – niagara Nov 26 '12 at 16:21
I assume you mean $\forall N \in \mathbb{N}, \int_1^N f(x)dx < \infty$, but this does not mean that $a_N$ is bounded (consider for instance the function $f(x) = 1$). – beauby Nov 26 '12 at 16:24
Sorry, I made this question a little shorter than original and while doing that I omitted some information. Now I correct it. – niagara Nov 26 '12 at 16:28

1 Answer 1

up vote 0 down vote accepted

Since $f$ is non-negative, $g(y) = \int_1^y f(x) dx$ is non-decreasing. Moreover, $\forall y \geq 1, a_{\lfloor y \rfloor} \leq g(y) \leq a_{\lceil y \rceil}$, so $g$ converges in $\infty$, which means that $\int_{1}^\infty f(x) dx$ converges (and $\lim_{y \to \infty} g(y) = \lim_{N \to \infty} a_N$).

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Thanks, but maybe that $f(y)$ in $a_{\lfloor y \rfloor} \leq f(y) \leq a_{\lceil y \rceil}$ means $g(y)$? – niagara Nov 26 '12 at 16:38
Yes, sorry. I edited it. – beauby Nov 26 '12 at 17:17

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