How to calculate the limit $\lim_{x \to 1} \frac{x + x^2 +\cdots+ x^n - n}{x - 1}$?

Hint. Note that $$\frac{x + x^2 +\dots+ x^n - n}{x - 1}=\frac{(x-1) + (x^2-1) +\dots+ (x^n - 1)}{x - 1}.$$ Moreover, for $k\geq 1$, $$\lim_{x\to 1}\frac{x^k-1}{x-1}=\lim_{x\to 1}\left(x^{k-1}+x^{k-2}+\dots +x+1\right)=k.$$

Hint:

L'Hospital's rule works well here.

Remember that $$\sum_{i=1}^n x^i=\frac{x \left(x^n-1\right)}{x-1}$$ Let $x=y+1$ and consider $$S_n=\frac {-n+\sum_{i=1}^n x^i }{x-1}=\frac{\frac{(y+1) \left((y+1)^n-1\right)}{y}-n}{y}$$ Now, using the binomial theorem or Taylor series $$(y+1)^n=1+n y+\frac{1}{2} (n-1) n y^2+\frac{1}{6} (n-2) (n-1) n y^3+O\left(y^4\right)$$ $$(y+1)^n-1=n y+\frac{1}{2} (n-1) n y^2+\frac{1}{6} (n-2) (n-1) n y^3+O\left(y^4\right)$$ $$\frac{(y+1) ^n-1}{y}=n +\frac{1}{2} (n-1) n y+\frac{1}{6} (n-2) (n-1) n y^2+O\left(y^3\right)$$ $$(y+1)\frac{(y+1) ^n-1}{y}=n+\frac{1}{2} n (n+1) y+\frac{1}{6} n \left(n^2-1\right) y^2+O\left(y^3\right)$$ $$(y+1)\frac{(y+1) ^n-1}{y}-n=\frac{1}{2} n (n+1) y+\frac{1}{6} n \left(n^2-1\right) y^2+O\left(y^3\right)$$ making $$S_n=\frac{1}{2} n (n+1) +\frac{1}{6} n \left(n^2-1\right) y+O\left(y^2\right)$$ which shows the limit and how it is approached when $y\to 0$.