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I aim to show that $\{a_n\} = \frac{1+nx^2}{(1+x^2)^n}$ converges to $0$. The following two facts seem obvious:

(1) $\forall n \in \mathbb{N}, a_n \ge 0$ (since each $a_n$ is the product of two positive numbers).

(2) $\forall n \in \mathbb{N}, a_n \ge a_{n+1}$.

Yet this is not quite enough to show that $a_n \rightarrow 0$ since the sequence could very well converge to some other $c > 0$.

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Presumably, you aim to show that for all $x \neq 0$, $a_n \to 0$ as $n \to \infty$. –  JavaMan Nov 4 '12 at 21:13

4 Answers 4

up vote 4 down vote accepted

Let $x\ne 0$. We will show that $$\lim_{w\to\infty}\frac{1+wx^2}{(1+x^2)^w}=0.$$ Use L'Hospital's Rule. It is clear that top and bottom blow up as $w\to\infty$. It follows that $$\lim_{w\to\infty}\frac{1+wx^2}{(1+x^2)^w}=\lim_{w\to\infty} \frac{x^2}{w(1+x^2)^{w-1}}.$$ Whenever $w\ge 1$, we have $w(1+x^2)^{w-1}\ge w$. Thus $$\lim_{w\to\infty} \frac{x^2}{w(1+x^2)^{w-1}}=0.$$

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The claim is not entirely true; it is valid if and only if $ x \in \mathbb{R} \setminus \{ 0 \} $. Here is a solution that uses neither transcendental functions nor calculus techniques such as l'Hôpital's Rule.

Firstly, observe that the limit is $ 1 $ if $ x = 0 $. Next, let $ n \in \mathbb{N} $. By the Binomial Theorem, we have $$ \forall x \in \mathbb{R}: \quad (1 + x^{2})^{n} \geq 1 + n x^{2} + \frac{n(n - 1)}{2} x^{4}. $$ Hence, \begin{align} \forall x \in \mathbb{R}: \quad \frac{(1 + x^{2})^{n}}{1 + n x^{2}} &\geq \frac{1 + n x^{2} + \frac{n(n - 1)}{2} x^{4}}{1 + n x^{2}} \\\\\\ &= \frac{1 + n x^{2}}{1 + n x^{2}} + \frac{\frac{n(n - 1)}{2} x^{4}}{1 + n x^{2}} \\\\\\ &= 1 + \frac{\frac{n(n - 1)}{2} x^{4}}{1 + n x^{2}} \\\\\\ &= 1 + \frac{\frac{1}{n} \left[ \frac{n(n - 1)}{2} x^{4} \right]}{\frac{1}{n} \left( 1 + n x^{2} \right)} \\\\\\ &= 1 + \frac{\frac{n - 1}{2} x^{4}}{\frac{1}{n} + x^{2}} \\\\\\ &= 1 + \frac{n - 1}{2} \cdot \frac{x^{4}}{\frac{1}{n} + x^{2}} \\\\\\ &\geq 1 + \frac{n - 1}{2} \cdot \frac{x^{4}}{1 + x^{2}}. \end{align} If $ x \neq 0 $, then the expression in the last line diverges to $ \infty $ as $ n \rightarrow \infty $. It follows that $$ \forall x \in \mathbb{R} \setminus \{ 0 \}: \quad \lim_{n \rightarrow \infty} \frac{(1 + x^{2})^{n}}{1 + n x^{2}} = \infty. $$ The original limit must therefore be $ 0 $.

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The problem is false for $x=0$. If $x \neq 0$, then

$$\lim_n \frac{1+nx^2}{(1+x^2)^n} = \lim_n \frac{1}{(1+x^2)^n}+x^2\lim_n \frac{n}{(1+x^2)^n}$$

It is trivial to show that

$$\lim_n \frac{1}{(1+x^2)^n} =0$$

While, for $x\neq 0$ we have

$$\lim_n \frac{n}{(1+x^2)^n} = \lim_n \left( \frac{\sqrt[n]{n}}{(1+x^2)} \right)^n=0$$

the last equality following from the fact that

$$ \lim_n \frac{\sqrt[n]{n}}{(1+x^2)} = \frac{1}{x^2+1} <1$$

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+1. Small typo, however: $\sqrt{n}$ should be $\sqrt[n]{n}$. –  JavaMan Nov 4 '12 at 21:26
    
@JavaMan Ty, fixed. Hopefully that is the only one :) –  N. S. Nov 4 '12 at 21:30
    
+1. Very good for getting the special case $x=0$. –  Mhenni Benghorbal Nov 4 '12 at 21:34

Note that $\log a_n=\log(x^2+1/n)+\log n-n\log(1+x^2)$. The first term converges to a finite limit and the second term is negligible compared to the third term since $\log(1+x^2)\ne0$. Furthermore, $\log(1+x^2)\gt0$, thus $\log a_n\to-\infty$. This proves that $a_n\to0$.

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