Evaluar lim $_{n\rightarrow\infty}$ $\frac{1-2+3-4+5-...............+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}$

Question

<blockquote> <p>Evaluar %#% $ #%</p> </blockquote> <p><strong>Mi enfoque</strong></p> <p>Que $$\lim_{n\rightarrow\infty}\frac{1-2+3-4+5-\cdots+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}.$ y $A_{n}=1+3+5+\cdots\left(2n-1\right)$ y $B_{n}=2+4+6+\cdots\left(2n\right)$ y $A_n=\frac{n}{2}\left(1+2n-1\right)=n^{2}$. Por lo tanto, $B_{n}=\frac{n}{2}\left(2+2n\right)=n+n^{2}$ $ pero el libro menciona que la respuesta es $$\lim_{n\rightarrow\infty}\frac{A_{n}-B_{n}}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=\lim_{n\rightarrow\infty}\frac{-n}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=-\frac{1}{2}.$.</p>

Answer 1

One may write $$ 1-2+3-4+5-\cdots+\left(-2n\right)=\underbrace{-\left(2-1\right)}{\color{red}{-1}}\:\underbrace{-\left(4-3\right)}{\color{red}{-1}}-\cdots\underbrace{-\left(2n-(2n-1)\right)}_{\color{red}{-1}}=\color{red}{-n} $$ dando $$ \frac{1-2+3-4+5-\cdots+\left(-2n\right)} {\sqrt {n ^ {2} +1} + \sqrt {n ^ 1 {2}}} = \frac {-n} {n\sqrt {1 +1/n ^ {2}} + n\sqrt {1-1/n ^ {2}}} \to-\frac12 $$ as $n\to \infty$.

El resultado es correcto para mí.

Answer 2

utilizar ese % $ $$\sum_{i=1}^n(-1)^{i-1}(2i)=(n-1)^2-(n+1)$