# proof of integral test

Consider the function (see the definition of floor)

 $g(x)=a_{\lfloor x\rfloor}.$

Clearly for $x\in[n,n+1)$, being $f$ non increasing we have

 $g(x+1)=a_{n+1}=f(n+1)\leq f(x)\leq f(n)=a_{n}=g(x)$

hence

 $\int_{M}^{+\infty}g(x+1)\,dx=\int_{M+1}^{+\infty}g(x)\,dx\leq\int_{M}^{+\infty% }f(x)\leq\int_{M}^{+\infty}g(x)\,dx.$

Since the integral of $f$ and $g$ on $[M,M+1]$ is finite we notice that $f$ is integrable on $[M,+\infty)$ if and only if $g$ is integrable on $[M,+\infty)$.

On the other hand $g$ is locally constant so

 $\int_{n}^{n+1}g(x)\,dx=\int_{n}^{n+1}a_{n}\,dx=a_{n}$

and hence for all $N\in\mathbb{Z}$

 $\int_{N}^{+\infty}g(x)=\sum_{n=N}^{\infty}a_{n}$

that is $g$ is integrable on $[N,+\infty)$ if and only if $\sum_{n=N}^{\infty}a_{n}$ is convergent.

But, again, $\int_{M}^{N}g(x)\,dx$ is finite hence $g$ is integrable on $[M,+\infty)$ if and only if $g$ is integrable on $[N,+\infty)$ and also $\sum_{n=0}^{N}a_{n}$ is finite so $\sum_{n=0}^{\infty}a_{n}$ is convergent if and only if $\sum_{n=N}^{\infty}a_{n}$ is convergent.

Title proof of integral test ProofOfIntegralTest 2013-03-22 13:32:07 2013-03-22 13:32:07 paolini (1187) paolini (1187) 4 paolini (1187) Proof msc 40A05