Let $u_n=f(n)$ for $n\geq 0$, $v_n=\int_{1}^nf(x)dx$ for $n\geq 1$. Show that the limit $$\lim_{n\to\infty}\left[ \left( \sum_{k=1}^n u_k\right) - v_n\right]$$ exists.
Rewriting the term in the limit this is
$$\sum_{k=1}^nf(k) - \int_1^n f(x)dx \leq nf(1) - \int_1^n f(x)dx$$ since $f$ is non-increasing. From here I am completely stuck and not sure if I am going in the correct direction. I'm certain I have to use properties of $f$ and its integral, but I cannot think of what to use. Any help would be much appreciated.
That's enough write \begin{equation} \int_1^n f(x)=\sum_{k=2}^n\int_{k-1}^kf(x) \end{equation} Now, using also that f is nonincreasing, the limit become \begin{equation} \left[ \left( \sum_{k=1}^n u_k\right) - v_n\right]=\sum_{k=2}^n\left(f(k-1)-\int_{k-1}^kf(x)\right)+f(n)\le \sum_{k=2}^n (f(k-1)-f(k))+f(n)=f(1) \end{equation} Hence the sequence $\{ \sum_{k=2}^n\left(f(k-1)-\int_{k-1}^kf(x)\right)+f(n)\}_{n\in Z_+}$ is bounded and, clearly increasing. Then the limit exist.