Find the limit of the sequence $\{x_n \}$.

Question

Let $\{x_n \}$ be a sequence defined by $x_n = \left(1 - \frac {1} {n^2} \right)^{\binom {n} {2}},\ n \geq 1$. Find the limit of the sequence $\{x_n \}$.

I find difficulty in doing this. Please help me in this regard.

Thank you very much.

Answer 1

$$ \left( \left( 1 - \frac{1}{n^2} \right)^{n^2} \right)^{\frac{n-1}{2n}} \to (e^{-1})^{\frac{1}{2}} = e^{-\frac{1}{2}} $$ This is true because, if you take the logarithm, you have $$ \log(x_n) = \binom{n}{2} \cdot \log(1- \frac{1}{n^2}) = \frac{n-1}{2n} \cdot \left( n^2 \cdot |log(1-\frac{1}{n^2}) \right)$$ The first part goes to $\frac{1}{2}$ and the second one goes to $-1$.

Thus their product goes to $-\frac{1}{2}$, hence the result.

Answer 2

$a_n= (1-1/n^2)^{n^2}$; $b_n = (1-1/n^2)^n;$

$x_n = (a_n/b_n)^{1/2}$;

$\lim_{n \rightarrow \infty}a_n = e^{-1}$.

(subsequence of $(1-1/n)^{1/n}$).

$b_n = [(1-1/n^2)^{n^2}]^{1/n}=$

$\exp ((1/n)\log ((1-1/n^2)^{n^2})$.

$\lim_{n \rightarrow \infty} b_n =e^0=1$.

Combining:

$\lim_{n \rightarrow \infty}x_n =$

$\lim_{n \rightarrow \infty}(a_n/b_n)^{1/2}= $

$[\frac{\lim_{n \rightarrow \infty}a_n}{\lim_{n \rightarrow \infty}b_n}]^{1/2}= e^{-1/2}$.

Answer 3

$$x_n = \left(1 - \frac {1} {n^2} \right)^{\binom {n} {2}}\implies \log(x_n)={\binom {n} {2}}\log\left(1 - \frac {1} {n^2} \right)\sim -\frac {\frac{1}{2} (n-1) n} {n^2}\sim - \frac{1}{2}$$