Limit of a polynomic-exponential sequence

Question

I have to calculate the following limit: $$L=\lim \limits_{n \to \infty} -(n-n^{n/(1+n)})$$ I get the indeterminate form $\infty - \infty$ and I don't know how to follow. Any idea? Thank you very much.

Answer 1

$$ n-n^{n/(1+n)} = n(1- e^{-1/(1+n) \log n})\sim n\times 1/(1+n) \log n \to\infty $$because when $u\to 0$,

$$ e^u-1\sim u $$

Answer 2

HINT : $$n - n^{n/1+n} = n - n^{1+1/n} = n(1 - n^{1/n}) = \frac{1 - n^{1/n}}{1/n}$$

Now use L'Hôpital's rule.

Answer 3

Since you received good answers, if I may, I suggest you plot on the same graph the two following functions $$f(n)= -(n-n^{n/(1+n)})$$ $$g(n)=-\log(n)$$ You will probably be amazed to see how close they are (even for $n$ around $100$).

In fact, but this is off topic, what you could show is that, for large values of $n$,$$f(n) \simeq \log \left(\frac{1}{n}\right)+\frac{\left(\log \left(\frac{1}{n}\right)-2\right) \log \left(\frac{1}{n}\right)}{2 n}+O\left(\left(\frac{1}{n}\right)^2\right)$$