<s>[INST] Show that $\sum_{n=1}^\infty a_n<\frac{1}{2}(f(1)-f(2))$. Let $f:[1,\infty)\rightarrow(0,\infty)$ be a twice differentiable decreasing function such that $f''(x)$ is positive for $x\in(1,\infty)$. For each positive integer $n$, let $a_n$ denote the area of the region bounded by the graph of $f$ and the line segment joining the points $(n,f(n)$ and $(n+1,f(n+1))$. Show that
$\sum_{n=1}^\infty a_n<\frac{1}{2}(f(1)-f(2))$.
My attempt: I can see that 
$a_n=f(n+1)+\frac{1}{2}(f(n)-f(n+1))-\int_n^{n+1}f(n)$
$=\frac{1}{2}(f(n)+f(n+1))-\int_n^{n+1}f(n)$
$<\frac{1}{2}(f(n)+f(n+1))-f(n+1)$
which is somewhat helpful in the infinite sum as many terms cancel, but I'm not getting the right bound. How can I use the second derivative being negative?
 [/INST] On each interval $[n,n+2]$, since $f$ is convex, $f$ is above its tangent line at $n+1$ (see property 5 here), hence for all $x\in [n,n+2]$, $$ f(x) \geq f(n+1)+ f'(n+1)(x-(n+1))$$
Integrating this inequality over $x$, 
$$\int_{n}^{n+2}f(x) dx \geq \int_{n}^{n+2} [f(n+1)+ f'(n+1)(x-(n+1))]dx$$
which rewrites $$\int_n^{n+2} f(x)dx \geq 2f(n+1)$$
Remember that $$a_n=\frac{1}{2}(f(n)+f(n+1))-\int_n^{n+1}f(n)$$
Hence $$\begin{align} a_n+a_{n+1} &= \frac{f(n)+f(n+1)}2 + \frac{f(n+1)+f(n+2)}2 - \int_{n}^{n+2} f(x)dx\\
&\leq \frac{f(n)+f(n+1)}2 + \frac{f(n+1)+f(n+2)}2 - 2f(n+1) \\
&=\frac{f(n)-f(n+1)}2 - \frac{f(n+1)-f(n+2)}2
\end{align} $$
Since $f$ is decreasing and positive, it has a finite limit at $\infty$ hence $\frac{f(n+1)-f(n+2)}2$ converges to $0$.
Summing all these inequalities, the sums on the right telescope, yielding $$\sum_{n=1}^\infty (a_n+a_{n+1})\leq \frac{f(1)-f(2)}2$$
that is $$\left( 2\sum_{n=1}^\infty a_n \right) -a_1 \leq \frac{f(1)-f(2)}2$$
$$\sum_{n=1}^\infty a_n \leq \frac{f(1)-f(2)}4 + \frac{a_1}2$$
The bound you found in your question yields $\displaystyle a_1\leq \frac{f(1)-f(2)}2$ and we're done.
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