Chaplin or Leibniz ?


Before I do proofs via the 3-step mathematical induction, the classic Charlie Chaplin movie clip often comes to my mind. It nudges me to seek better ways. For example, to prove

(k+1)(1^k+2^k+3^k+\dots+n^k)<(n+1)^{k+1}, \quad k, n \in N^+,

Instead of applying the 3-step mathematical induction, let us look at Fig. 1.

Screen Shot 2017-07-07 at 6.44.28 PM.png

Fig. 1

If A denotes the total area of the blue rectangles, Fig.1 shows

A=1\cdot 1^k+1\cdot 2^k+1\cdot 3^k+\dots+1\cdot n^k = 1^k+2^k+3^k+\dots+n^k.

It also shows that

A < the  area under the curve x^k = \int\limits_{0}^{n+1}x^k\; dx.

By the fundamental theorem of calculus,

\int\limits_{0}^{n+1}x^k\;dx = \frac{x^{k+1}}{k+1}\bigg|_{0}^{n+1}=\frac{(n+1)^{k+1}}{k+1}.


1^k+2^k+3^k+\dots+n^k < \frac{(n+1)^{k+1}}{k+1},


(k+1)(1^k+2^k+3^k+\dots+n^k)<(n+1)^{k+1}, \quad\quad k, n \in N^+.

This proof suggests another inequality:

(k+1)(1^k+2^k+3^k+\dots+n^k)>n^{k+1}, \quad\quad k, n \in N^+.

One thought on “Chaplin or Leibniz ?

  1. Pingback: “Chaplin or Leibniz ?” Revisit | Vroom

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