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?
 A: I am jumping directly to the step where you have reached.
$$a_n=f(n+1)+\frac{1}{2}(f(n)-f(n+1))-\int_n^{n+1}f(n)$$
Now $\int_n^{n+1}f(n)$ represents the area under the graph from $n$ to $n+1$.
If we replace it by a smaller quantity, then we will increase its value.
But from what value should we replace it? You have used $f(n+1)$ but it doesn't give required result .So let us use graphs.This is a random decreasing convex graph.

Here that blue line is tangent to the curve at point $B(n+1,f(n+1))$
Let us subtract the area enclosed by the curve $ABD$. 
Now, what we are left with is $\int_n^{n+1} \text{ Line} BD $.
On integrating the line
$y=f'(n+1)x+f(n+1)-f'(n+1)(n+1)$  
On $n$ to $n+1$, what we get is :
$$\int_n^{n+1}f'(n+1)x+f(n+1)-f'(n+1)(n+1)=f(n+1)-\frac {f'(n+1)}{2}$$ 
(Notice, f'(n+1) is negative here.)
Now we are done.
$$\implies a_n = \frac  {(f(n)+f(n+1))}{2}-\int_n^{n+1}f(n)$$
$$\implies a_n < \frac {1}{2}(f(n)+f(n+1))-(f(n+1)-\frac {f'(n+1)}{2})$$
$$ \implies a_n < \frac{1}{2}(f(n)-f(n+1))+\frac {f'(n+1)}{2}$$
$$\implies \sum_{n=1}^\infty a_n<\ \sum_{n=1}^\infty \Bigg[\frac{1}{2}(f(n)-f(n+1))+\frac {f'(n+1)}{2}\Bigg]$$
$$ \implies \sum_{n=1}^\infty a_n< \lim _{m \rightarrow \infty}\Bigg[ \frac{1}{2}(f(1)-f(m))+\ \sum_{n=1}^m \Bigg (\frac {f'(n+1)}{2}\Bigg )\Bigg]$$
Now, hold on for a minute. Notice:
$$\sum_{n=1}^\infty \Bigg (\frac {f'(n+1)}{2}\Bigg )$$ 
can be approximated as -
$$\int_1^{\infty} \frac {f'(n+1)}{2}$$
(using sum as integration since $1,2,3...$ are very small intervals as compared to $\infty$)
Continue :
$$\implies \sum_{n=1}^\infty a_n< \lim _{m \rightarrow \infty} \Bigg[\frac{1}{2}(f(1)-f(m))+\int_1^{m} \frac {f'(n+1)}{2} \Bigg]$$
$$\implies \sum_{n=1}^\infty a_n< \lim _{m \rightarrow \infty} \Bigg[\frac{1}{2}(f(1)-f(m))+ \frac {f(n+1)}{2} \Bigg |_{1}^{m} \Bigg ]$$
$$\implies \sum_{n=1}^\infty a_n<  \frac{1}{2}(f(1)-f(m))+ \frac {f(m)-f(1+1)}{2}$$
$$\implies \sum_{n=1}^\infty a_n<  \frac{1}{2}(f(1)-f(2))$$
Finally, done with it!
A: 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.
A: Consider the following figure, where the area $a_n$ is shaded in red:

Convexity implies that the area $a_n$ is contained in the triangle $\triangle ABC$. It follows that
$$\eqalign{a_n&\leq {\rm area}(\triangle ABC)={1\over2}\bigl(f(n)-2f(n+1)+f(n+2)\bigr)\cr&={1\over2}\bigl(f(n)-f(n+1)\bigr)-{1\over2}\bigl(f(n+1)-f(n+2)\bigr)\ .\cr}$$
Summing over $n$ gives on the right hand side a (convergent) telescopic sum, so that we obtain
$$\sum_{n=1}^\infty a_n\leq{1\over2}\bigl(f(1)-f(2)\bigr)\ .$$
