Proving increasing property of subseries

30 Views Asked by At

Plotting the series $x^n,x^{n+1},x^{n+2},\ldots,x^{z}$, for any $x,n,z\in\mathbb{N}$ I get evidence that each sum of consecutive terms is lower than the next consecutive term:

1, 2, 4, 8, 16, 32...

1, 3, 9, 27, 81, 243...

1, 10, 100, 1000, 10000, 10000...

Where, for example:

$1+2+4\lt 8$,

$3+9+27<81$,

etc.

But how can this actually be proved?

1

There are 1 best solutions below

0
On

The property doesn't hold for $x=0$ or $1$, so assume $x\geq2$.

Now for any natural number $a\leq b$, using the sum formula for geometric series we have

$$x^a+x^{a+1}+\cdots+x^b\leq 1+x+x^2+\cdots+x^b=\frac{x^{b+1}-1}{x-1}<x^{b+1},$$

where the last step is due to $x-1\geq1$ and $x^{b+1}-1<x^{b+1}$.