# Properties of a twice differentiable decreasing function $f$ such that $f''(x)$ is positive for $x \in (1, \infty$).

Let $f : [1,\infty) \to (0,\infty)$ 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 $(n + 1, f(n + 1))$. I want to show
a. $\displaystyle \sum_{n=1}^ \infty a_{n}<\frac 1 2 (f(1)-f(2))$

b. $\displaystyle\lim_{n \to \infty} \left[ \sum_{k=1}^nf(k)-\frac1 2(f(1)+f(n))- \int_1^n f(x)dx\right]$ exists.

c. $\displaystyle\lim_{n \to \infty} \left[ \sum_{k=1}^nf(k)\int_1^n f(x)dx\right]$ exists.

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What have you tried? –  user21436 Mar 10 '12 at 6:23
And, consider registering your account. This has several benefits in addition to helping system maintain your posts efficiently, and helping you edit your own posts. –  user21436 Mar 10 '12 at 6:25

• I don’t immediately see why (a) is true, though it’s quite easy to see that $$\sum_{n\ge 1}a_n<\frac12\Big(f(1)+f(2)\Big)$$ just by sliding each of the slivers over to the column between $x=1$ and $x=2$.

• For (b) note that $$\left(\sum_{k=1}^nf(k)-\frac12\big(f(1)+f(n)\big)\right)-\int_1^nf(x)\,dx=\sum_{k=1}^na_k\;:$$ the term in large parentheses on the left is the area of a bunch of trapezoids.

• (c) is false as stated: try it with $f(x)=\dfrac1x$.

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The Consider the function $f$ on the iterval $[n,n+1]$. Since $f$ is convex, the graph of $f$ is above the tangent at the point $(n+1,f(n+1))$ and below the secant passing through $(n,f(n))$ and $(n+1,f(n+1))$. See the image below.
Then $a_n$ is less than the area of the triangle formed by the red lines (secant and tangent) and the vertical line $x=n$. Thus $$a_n\le \frac12\,(f(n)-f(n+1)+f'(n+1)).$$ Then \begin{align*} \sum_{k=1}^na_k&\le\frac12\,\Bigl(f(1)-f(2)+f(2)-f(3)+\dots+f(n)-f(n+1)+\sum_{k=2}^{n+1}f'(k)\Bigr)\\ &=\frac12\,\Bigl(f(1)-f(n+1)+\sum_{k=2}^{n+1}f'(k)\Bigr) \end{align*} But $$f(n+1)-f(2)=\int_2^{n+1}f'(x)\,dx=\sum_{k=2}^{n}\int_k^{k+1}f'(x)\,dx\ge\sum_{k=2}^{n}f'(k),$$ since $f'$ is increasing. It folllows that $$\sum_{k=1}^na_k\le\frac12(f(1)-f(2)+f'(n+1))\le\frac12(f(1)-f(2)),$$ proving a. You should be able to do b. and c. (where I think there is a $-$ sign missing.) If you have any trouble, search for Euler's summation formula.