The question is, if $f:[1,\infty)\rightarrow [0,1)$ a non-increasing function with $\lim_{x\to \infty}f(x)=0$. Then Show that $\lim_{n\to \infty}[\int_1^2 f^n(x)+\int_2^3 f^{n-1}(x)+......+ \int_{n-1}^n f(x)]=0.$
Proof: Since $\lim_{x\to \infty}f(x)=0$ then we have For a fixed $\varepsilon>0,\exists$ a sufficiently large $N \in\Bbb N$, such that $|f(x)|<\frac{\varepsilon}{2N+1}$ $\forall x>N$. $$0<f(x)≤M<1,M\in\Bbb R \implies f^n(x)≤M^n<\frac{\varepsilon}{2N+1}\forall n>N$$ for sufficiently large $N$. $\implies \int_i^{i+1} f^n(x)<\frac{\varepsilon}{2N+1}$
Now for $n=2N$ we have, $$\sum_{i=1}^{2N-1}\int_i^{i+1}f^{n+1-i}(x)dx= \sum_{i=1}^{N+1}\int_i^{i+1}f^{n+1-i}(x)dx+\sum_{i=N+2}^{2N-1}\int_i^{i+1}f^{n+1-i}(x)dx< \frac{(N+1)\varepsilon}{2N+1}+ \frac{(N-2)\varepsilon}{2N+1}=\frac{(2N-1)\varepsilon}{2N+1}<\varepsilon $$ This is the desired result.
I'm little bit confused about my proof. Somebody please help. Also tell me if there needs more improvement/correction to make this proof better. Thanks in advance.
Edited: because the Codomain was wrong, now Corrected.