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Suppose $f(x)$ is a decreasing function for all $x$ , and that $f(x) \geq 0$ for all $x$. Then,

$$\sum_{k=1}^N f(k) \geq \int_{1}^{N+1} f(x)\hspace{1mm} dx$$

Proving this result using a simple sketch is quite easy, as the sum will essentially act as an overestimated Riemann sum starting at $k=1$ and "reaches" up to $N+1$. However, I'm curious if anyone has a slightly more analytical approach to this proof, as I appriciate such proofs more.

William
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1 Answers1

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Since $f$ is decreasing, $f(k) = \int_k^{k+1} f(k)\, dx \ge \int_k^{k+1} f(x)\, dx$ and hence $$\sum_{k = 1}^N f(k) \ge \sum_{k = 1}^N \int_k^{k+1} f(x)\, dx = \int_1^{N+1} f(x)\, dx$$

kobe
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  • Could you further elaborate why it is that $f(k) \geq F(k+1)- F(k)$ ? – William May 11 '22 at 15:51
  • @William if $x\ge k$ then $f(k) \ge f(x)$ by the decreasing assumption on $f$. So then $\int_k^{k+1} f(k) dx \ge \int_k^{k+1} f(x), dx$. – kobe May 11 '22 at 15:53
  • Doesn’t the proof always then rely on $x \geq k$? – William May 11 '22 at 17:04
  • @William I don't quite understand the question. For all $x\in [k,k+1]$ we have $f(k) \ge f(x)$; we then integrate both sides of this inequality from $x = k$ to $x = k+1$ to get $f(k) \ge \int_k^{k+1} f(x), dx$. – kobe May 11 '22 at 17:07
  • Thank you, this clarified it for me! – William May 11 '22 at 18:44