Limit of the $n^{2}\left( \ln\left( 1+\frac{1}{n}\right) -\frac{1}{n+1}\right)$

Question

How can one compute the following limit $$ \lim_{n\rightarrow+\infty}n^{2}\left( \ln\left( 1+\frac{1}{n}\right) -\frac{1}{n+1}\right) \quad ? $$ I wrote

$$ \lim_{n\rightarrow+\infty}n^{2}\left( \ln\left( 1+\frac{1}{n}\right) -\frac{1}{n+1}\right) =\lim_{x\rightarrow0}\frac{\ln\left( 1+x\right) -x}{x^{2}}\overset{l^{\prime}H}{=}\lim_{x\rightarrow0}\frac{\dfrac{1}{1+x}% -1}{2x}\overset{l^{\prime}H}{=}\lim_{x\rightarrow0}\frac{-1}{2\left( 1+x\right) ^{2}}=-\frac{1}{2}. $$ But how can be a limit of a positive sequence be negative?

Answer 1

$$ \lim_{n\rightarrow+\infty}n^{2}\left( \ln\left( 1+\frac{1}{n}\right) -\frac{1}{n+1}\right) $$ $$ =\lim_{n\rightarrow+\infty} \left( n^2 \ln\left( 1+\frac{1}{n}\right) -n\frac{1}{1+\frac1n}\right) $$

$$ =\lim_{n\rightarrow+\infty} \left(n^2\left[ \dfrac1n-\dfrac1{2n^2}+O\left(\dfrac1{n^3}\right)\right]- n\left[1-\dfrac1n+O\left(\dfrac1n\right)\right]\right)= $$

$$ =\lim_{n\rightarrow+\infty} \left( \left[ n-\dfrac1{2 }+O\left(\dfrac1{n}\right)\right]- \left[n-1+O\left(\dfrac1n\right)\right]\right)=\dfrac12 $$

Answer 2

Let $\dfrac1n=h,\dfrac1{1+n}=\dfrac1{1+\dfrac1h}=\dfrac h{h+1}$

$$\lim_{h\to0}\dfrac{\ln(1+h)-\dfrac h{h+1}}{h^2}$$

$$=\lim_{h\to0}\dfrac{\ln(1+h)-h}{h^2}+\lim_{h\to0}\dfrac{h-\dfrac h{h+1}}{h^2}$$

Use Are all limits solvable without L'Hôpital Rule or Series Expansion to find the first limit $=-\dfrac12$

For the second, its $=1$ right?