This is exercise 10.24.20 in Apostol, Calculus 1. The question has been asked before (Does the existence of $\lim_{n \to \infty}\int_1^n f(x)dx $ imply convergence of$\int_1^\infty f(x)dx$ if $f$ is decreasing?) but I don't think it is resolved there.
The following is my proof. Could you please check if it is correct.
We first prove that $f(x)$ has a fixed sign for all $x\geq c$ for some $c>1$. If there exists a $a>1$ such that $f(a)<0$, then $f(x)\leq f(a)<0$ for all $x\geq a$. Otherwise, assume no such $a$ exists. If there exists a $b>1$ such that $f(b)=0$, then $f(x)\leq f(b)=0$ for all $x\geq b$. But $f(x)$ cannot be negative (contradicting the hypothesis), we have $f(x)=0$ for all $x>b$. If there exists no such $b$, then $f(x)>0$ for all $x\geq 1$. Therefore there exists a positive integer $m\geq 1$, such that $x\geq m$ implies $f(x)$ has a fixed sign.
We now consider the three cases: (i) If $f(x)=0$ for $x\geq m$, then $\int_1^\infty f(x)dx$ converges, since $\int_m^\infty f(x)dx=0$.
(ii) If $f(x)<0$ for $x\geq m$, then $f(x)\leq f(m)<0$. For $n\geq m$, $I_n-I_m=\int_m^n f(x)dx\leq\int_m^n f(m)dx=(n-m)f(m)$ does not have a limit as $n\to+\infty$, contradicting that $\{I_n\}$ converges.
(iii) The only remaining possibility is $f(x)>0$ for all $x>0$. $f$ is monotonic decreasing with a lower bound, so $L=\lim_{x\to+\infty} f(x)$ exists. If $L>0$, then $I_n=\int_1^n f(x)dx>\int_1^n Ldx=L(n-1)$ does not have a limit as $n\to\infty$, contradicting that $\{I_n\}$ converges. Therefore $\lim_{x\to+\infty} f(x)=0$. Thus $\int_1^\infty f(x)dx$ converges, by exercise 10.24.21.
The content of exercise 10.24.21 is: If $\lim_{x\to\infty} f(x)=0$ and $\lim_{n\to\infty} I_n=A$, then $\int_1^\infty f(x)dx$ converges and has the value $A$.
Proof of exercise 10.24.21: Let $\epsilon>0$ be arbitrary. Since $\lim_{n\to\infty} I_n=A$, we could find a positive integer $N$ such that $n\geq N$ implies $|I_n-A|<\frac{\epsilon}{2}$. On the other hand, $\lim_{x\to\infty} f(x)=0$, so there exists a real number $M>0$ such that $x>M$ implies $|f(x)|<\frac{\epsilon}{2}$. Now for all $x>\max(M,N)+1$, we have $[x]\geq M,N$, so $|I_{[x]}-A|<\frac{\epsilon}{2}$ and $|f(x)|<\frac{\epsilon}{2}$, so $|\int_1^x f(t)dt-A|=|I_{[x]}+\int_{[x]}^x f(t)dt-A|\leq|I_{[x]}-A|+|\int_{[x]}^x f(t)dt|\leq|I_{[x]}-A|+\int_{[x]}^x |f(t)|dt<\frac{\epsilon}{2}+(x-[x])\frac{\epsilon}{2}<\epsilon$ (since $x-[x]<1$). Therefore $\lim_{x\to+\infty}\int_1^x f(t)dt=A$. Hence $\int_1^\infty f(x)dx$ converges and has the value $A$.
$[x]$ denotes the greatest integer satisfying $\leq x$.