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If [tex]$x\ \textless \ 1$[/tex], then

[tex]\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots = \infty[/tex]

A. [tex]1-x[/tex]
B. [tex]\frac{1}{x-1}[/tex]
C. [tex]\frac{1}{1+x}[/tex]
D. [tex]1+x[/tex]


Sagot :

Given the series:
[tex]\[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots \][/tex]

We aim to find the sum of this infinite series for [tex]\(x < 1\)[/tex].

First, let's rewrite the general term of the series:
[tex]\[ a_n = \frac{2^n x^{2^n - 1}}{1 + x^{2^n}} \][/tex]

To solve this, we observe the pattern and behavior of the terms in the series. The series appears to combine terms with increasing powers of [tex]\(x\)[/tex] in both the numerator and denominator as follows:

- The first term is [tex]\(\frac{1}{1+x}\)[/tex].
- The second term is [tex]\(\frac{2x}{1+x^2}\)[/tex].
- The third term is [tex]\(\frac{4x^3}{1+x^4}\)[/tex], and so forth.

Careful analysis and observation allow us to conjecture that the given series simplifies remarkably to a simple closed form. We hypothesize that all these terms sum up to [tex]\(1 - x\)[/tex] for [tex]\(x < 1\)[/tex].

Therefore, the sum of the series:
[tex]\[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots \][/tex]

is [tex]\(\boxed{1 - x}\)[/tex].