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Special Integrals - 1

If n is a p...

Question

If $n$ is a positive integer and $x < 1$ , show that
$\frac{1 - x^{n + 1}}{n + 1} < \frac{1 - x^{n}}{n}$ .

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Solution

We can write $\frac{1 - x^{n + 1}}{1 - x^{n}} = 1 + \frac{x^{n} (1 - x)}{1 - x^{n}}$

$\frac{1 - x}{1 - x^{n}} = \frac{1}{{(1 + x)}^{n} - x^{n}} = \frac{1}{1 + x + x^{2} + . . + x^{n - 1}}$

$\frac{1 - x^{n + 1}}{1 - x^{n}} = 1 + \frac{x^{n}}{1 + x + x^{2} + . . + x^{n - 1}}$

$\frac{1 - x^{n + 1}}{1 - x^{n}} = 1 + \frac{1}{\frac{1}{x^{n}} + \frac{1}{x^{n - 1}} + . . . + \frac{1}{x}}$

As $x < 1$

$\frac{1}{x} > 1$

$\frac{1}{x}, \frac{1}{x^{2}}, . ., \frac{1}{x^{n}}$ is greater than $1$ , and hence their sum is greater than $n$

$\frac{1}{x} + \frac{1}{x^{2}} + . . . + \frac{1}{x^{n}} > n$

$1 + \frac{1}{\frac{1}{x} + \frac{1}{x^{2}} + . . . + \frac{1}{x^{n}}} < 1 + \frac{1}{n}$

$\frac{1 - x^{n + 1}}{1 - x^{n}} < \frac{1 + n}{n}$

$\frac{1 - x^{n + 1}}{n + 1} < \frac{1 - x^{n}}{n}$

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