Improper integrals and sequences

woo · **Joined:** Fri, 13 Dec 2013 14:47:37 UTC **Posts:** 200

Let

be a continuous, positive, decreasing function deﬁned on $[1, \infty)$ . How to show that the improper integral $\int_{1}^{\infty} f(x) \,dx$ converges iff the sequence $\{a_n\}$ defined by $a_n=\int_{1}^{n} f(x) \,dx$ converges by using only theorems/facts from calculus I and II? Thank you.

Can we prove it by using the inequality $a_n\leq\int_{1}^{t} f(x) \,dx\leq a_{n+1}$ for $t\in[n,n+1]$ ? (The function $\int_{1}^{t} f(x) \,dx$ is continuous and increasing and the sequence ${\int_{1}^{n} f(x) \,dx}$ is increasing.) How can I show that the function $\int_{1}^{t} f(x) \,dx$ is bounded above if $\int_{1}^{\infty} f(x) \,dx$ is convergent? If I can show that the function $\int_{1}^{t} f(x) \,dx$ is bounded above then I can show that the sequence ${\int_{1}^{n} f(x) \,dx}$ is convergent.

S.O.S. Mathematics CyberBoard

Improper integrals and sequences

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