Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

Some hints and comments:

  • 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 $f(x)=\dfrac1x$.

share|cite|improve this answer

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.

enter image description here

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.