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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?

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

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  • $\begingroup$ I like this solution. Thanks! $\endgroup$ – Jason Apr 24 '18 at 11:19
  • $\begingroup$ You are welcome. Good luck. $\endgroup$ – farruhota Apr 24 '18 at 12:06
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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 $a_n=e^{\log(a_n)}$.

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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 $\lim_{x\to 0} \ln(1+x)/x=1$.

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  • $\begingroup$ Thanks, this seems to be a clean solution but I don't see where the negative sign came from. Also could you please elaborate a tiny bit more on the explanation of the step with the $\ln$ limit? $\endgroup$ – Jason Apr 24 '18 at 10:55
  • $\begingroup$ @Jason The negative sign comes from the logarithm, as $\ln(n/(n+3)) = -\ln((n+3)/n)$. $\endgroup$ – Hanul Jeon Apr 24 '18 at 10:57
  • $\begingroup$ @Jason The limit for logarithm is well-known. It is in fact a derivative of $f(x)=\ln(1+x)$ at 0, $\endgroup$ – Hanul Jeon Apr 24 '18 at 10:58
  • $\begingroup$ I understand the limit of $\ln(1+x)/x=1$ but what I'm missing is how you used it in this example $\endgroup$ – Jason Apr 24 '18 at 11:01
  • $\begingroup$ @Jason Replace $x$ to $3/n$, then you may see how it is used. $\endgroup$ – Hanul Jeon Apr 24 '18 at 11:01

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