Show that $\{a_n\}$ is non-increasing, non-negative and $0< \lim_{n \to \infty} Suppose that $f$ is positive, decreasing and locally integrable on $[1,\infty]$, and let
$$a_n = \sum_{k=1}^n f(k)-\int_1^n f(x)dx$$
Show that $\{a_n\}$ is non-increasing, non-negative and 
$$0< \lim_{n \to \infty} <f(1)$$
My solution:
We observe that 
$$0<nf(n)<\sum_{k=1}^n f(k) < nf(1) \text{ and } 0<(n-1)f(n)<\int_1^n f(x)dx < (n-1)f(1)$$
Therefore
$$0<f(n)<\sum_{k=1}^n f(k)-\int_1^n f(x)dx<f(1)$$
We now show that $\{a_n\}$ is non-increasing. Consider
$$a_{n+1}-a_n=f(n+1)-\int_n^{n+1}f(x)dx \leq f(n+1)-f(n+1)=0$$
Thus, our sequence is non-increasing.
My problem:
I have got the limit bound such that
$$0 \leq \lim a_n \leq f(1)$$
But I haven't been unable to get the inequalities to be strict. How do I get this done? I've explored what happens when the limit is $0$ or $f(1)$ but I haven't gotten anything productive out of it yet. Thanks for your help!             
 A: No.
You can't look at the whole
integral and sum.
To see why, note that
if
$a < b < c$
and
$d < e < f$,
then it is not true that
$a-d < b-e < c-f$.
You can not subtract inequalities,
you can only add them.
By rewriting the
second inequality as
$-f < -e < -d$,
all you can deduce is that
$a-f < b-e < c-d$.
What you need to do
is compare
$f(k)$
with
$\int_k^{k+1} f(x) dx$
and add the results.
Since $f$ is decreasing,
$f(k)
>\int_k^{k+1} f(x) dx
> f(k+1)
$.
Summing these,
$\sum_{k=1}^{n-1}f(k)
>\sum_{k=1}^{n-1}\int_k^{k+1} f(x) dx
=\int_1^{n} f(x) dx
$
and
$\sum_{k=1}^{n-1}f(k+1)
<\sum_{k=1}^{n-1}\int_k^{k+1} f(x) dx
=\int_1^{n} f(x) dx
$
so
$\sum_{k=2}^{n}f(k)
<\int_1^{n} f(x) dx
$.
Therefore
$\int_1^{n} f(x) dx
< \sum_{k=1}^{n-1}f(k)
= \sum_{k=1}^{n}f(k) - f(n)
$
and
$\int_1^{n} f(x) dx
>\sum_{k=2}^{n}f(k)
=\sum_{k=1}^{n}f(k) -f(1)
$
so that
$f(n)
< \sum_{k=1}^{n}f(k)-\int_1^{n} f(x) dx
< f(1)
$.
A: Since $f$ is positive and decreasing,
$$f(n+1) < \int_n^{n+1} f(x) \, dx = \int_1^{n+1} f(x) \, dx - \int_1^{n} f(x) \, dx < f(n).$$
Hence,
$$f(n+1) -  \int_1^{n+1} f(x) \, dx < - \int_1^{n} f(x) \, dx,$$
and the sequence  $a_n$ is strictly decreasing since
$$a_{n+1}= \sum_{k=1}^{n+1}f(k) -  \int_1^{n+1} f(x) \, dx < \sum_{k=1}^{n}f(k)- \int_1^{n} f(x) \, dx = a_n.$$
In particular,
$$f(1) > \int_1^2 f(x) \, dx > f(2),$$
and
$$f(2) = f(1) + f(2) - f(1) < f(1) + f(2) - \int_1^2f(x) \, dx < f(1) + f(2) - f(2) = f(1).$$
Therefore, for all $n > 2$ we have
$$a_n < a_2 = f(1) +f(2) - \int_1^2f(x) \, dx < f(1),$$
and
$$\lim_{n \to \infty} a_n \leqslant a_2 < f(1).$$
On the other hand,
$$\sum_{k=2}^{n}f(k) > \sum_{k=2}^{n-1}f(k) > \int_2^{n} f(x) \, dx, $$
and
$$\sum_{k=1}^{n}f(k) - f(1) >  \int_1^{n} f(x) \, dx -   \int_1^{2} f(x) \, dx. $$
Therefore, for all $n > 2$ we have
$$a_n = \sum_{k=1}^{n}f(k) - \int_1^{n} f(x) \, dx > f(1) -  \int_1^{2} f(x) \, dx > 0,$$
and
$$\lim_{n \to \infty} a_n \geqslant f(1) -  \int_1^{2} f(x) \, dx > 0.$$
