Calculate the following series limit: $\lim_{n \to \infty} (\frac{n}{n+3})^\sqrt{n(1+n)}$

Question

Calculate the following series limit: $$\lim_{n \to \infty} \left(\frac{n}{n+3}\right)^\sqrt{n(1+n)}$$

I'm struggling with this limit problem. I changed over to this form: $e^{\sqrt{n(1+n)}\ln(\frac{n}{n+3})}$ but I'm not sure how to continue from here, tried using Lhopital but it just ended up being nasty maybe I used it too early?

Answer 1

It is: $$\lim_{n \to \infty} \left(\frac{n}{n+3}\right)^\sqrt{n(1+n)}=\lim_{n \to \infty} \left(1+\frac{3}{n}\right)^{-\sqrt{n(1+n)}}=\lim_{n \to \infty} \left[\left(1+\frac{3}{n}\right)^{\frac{n}{3}}\right]^{-\frac{3\sqrt{n(1+n)}}{n}}=e^{-3}.$$

Answer 2

Hint $$-\lim_{n\to\infty}\sqrt{n(1+n)}\ln\left(1+\frac{3}{n}\right) = -\lim_{n\to\infty}\sqrt{n(1+n)}\cdot\frac{3}{n}$$ as $\lim_{x\to 0} \ln(1+x)/x=1$.

Answer 3

Consider $$a_n=\left(\frac{n}{n+3}\right)^{\sqrt{n (n+1)}}\implies \log(a_n)={\sqrt{n (n+1)}}\log\left(1-\frac3 {n+3} \right)$$ Using equivalents for large values of $n$ $${\sqrt{n (n+1)}}\sim n \qquad \text{and} \qquad \log\left(1-\frac3 {n+3} \right)\sim -\frac3 {n+3}\sim -\frac 3n$$ making $\log(a_n) \sim -3\implies a_n \sim e^{-3}$.

If you want to go further, use Taylor series for $\log(a_n)$ and continue with them using $a_n=e^{\log(a_n)}$.