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$\lim_{n\rightarrow \infty} \frac{x^n}{1+x^n}$ converges pointwise to

$0$, if $0≤x<1$.
$\frac{1}{2}$, if $x=1$.
$1$, if $x>1$

Which is seen by checking the conditions first for $\lim_{n\rightarrow \infty} x_n$ and then for $\lim_{n\rightarrow \infty} \frac{x_n}{1+x_n}$.

I'd like to understand how the case for $x>1$ is seen.

mavavilj
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    Why do you think the sum is $0$ if $0 < x < 1$? The sum of positive series is positive. – Umberto P. Nov 12 '15 at 14:07
  • Do you mean the sum, or just the terms in the sum? – MPW Nov 12 '15 at 14:08
  • These statements are true for the $\lim_{n \to \infty}{\frac{x^n}{1+x^n}}$ but not for the $\sum_{n =0}^{\infty}{\frac{x^n}{1+x^n}}$. – mzp Nov 12 '15 at 14:14

1 Answers1

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Note: $$\frac{1}{x^n}>\frac{1}{x^n+1}>0$$

so:

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

So if $x>1$ then $\frac{1}{x^n}\to 0$, and apply the squeeze theorem.

Thomas Andrews
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