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

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

hence

$${\int }_M^{+\mathrm{\infty }}g(x+1)𝑑x={\int }_{M+1}^{+\mathrm{\infty }}g(x)𝑑x\le {\int }_M^{+\mathrm{\infty }}f(x)\le {\int }_M^{+\mathrm{\infty }}g(x)𝑑x.$$

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

On the other hand $g$ is locally constant so

$${\int }_n^{n+1}g(x)𝑑x={\int }_n^{n+1}{a}_n𝑑x=a_n$$

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

$${\int }_N^{+\mathrm{\infty }}g(x)=\sum _{n=N}^{\mathrm{\infty }}{a}_n$$

that is $g$ is integrable on $[N,+\mathrm{\infty })$ if and only if ${\sum }_{n=N}^{\mathrm{\infty }}{a}_n$ is convergent.

But, again, ${\int }_M^{N}g(x)𝑑x$ is finite hence $g$ is integrable on $[M,+\mathrm{\infty })$ if and only if $g$ is integrable on $[N,+\mathrm{\infty })$ and also ${\sum }_{n=0}^N{a}_n$ is finite so ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n$ is convergent if and only if ${\sum }_{n=N}^{\mathrm{\infty }}{a}_n$ is convergent.