Prove a limit avoiding differentiation

Question

Is there a way to prove the following limit without differentiation? (Taylor, l'Hôpital, ...) $$ \lim_{n\to\infty} n\left(\left(\frac{n}{n+1}\right)^{1+\tfrac{c}{2}}-1\right) =-1-\tfrac{c}{2},\quad c>0 $$ Equivalently $$ \lim_{n\to\infty} n\left(\left(\frac{n}{n+1}\right)^{a}-1\right) =-a,\quad a>1 $$

Answer 1

The expression under limit can be written as $$\frac{\exp(-a\log(1+t))-1}{t}$$ where $t=1/n$. A little algebra shows that the above can be written as $$\frac{\exp(-a\log(1+t))-1}{-a\log(1+t)}\cdot\frac{\log(1+t)}{t}\cdot(-a)$$ which clearly tends to $-a$ as $t\to 0$. The result holds for any $a$ and there is no need of restriction $a>1$.

Answer 2

Write this as $$\lim_{n\to\infty} n\left(\left(1-\frac{1}{n+1}\right)^{a}-1\right) =-a,\quad a>1$$ and then use a binomial expansion of $\left(1-\frac{1}{n+1}\right)^{a}= 1 - \frac{a}{n+1}+O\left(\frac{1}{(n+1)^2}\right)$ to get $$\lim_{n\to\infty} -\frac{an}{n+1}+O\left(\frac{n}{(n+1)^2}\right) = -a$$