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

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.