## Limit of the sum of a series around 1

For a fixed $$n\in\mathbb N$$ compute the limit $$\displaystyle \lim_{x\to 1}\frac{x+x^2+…+x^n-n}{x-1}$$

• #### Resolution

We can use the theorem about the limit of a sum $$n$$ times and we obtain

$$\lim_{x\to 1}\frac{x+x^2+…+x^n-n}{x-1} =\lim_{x\to 1}\frac{(x-1)+(x^2-1)+…+(x^n-1)}{x-1}\\ = \lim_{x\to 1}\frac{x-1}{x-1}+ \lim_{x\to 1}\frac{x^2-1}{x-1}+…+\lim_{x\to 1}\frac{x^n-1}{x-1}$$.

For $$j\in\{1{,}2,…,n\}$$ we can use the factorization of $$a^j-b^j$$ to get

$$\displaystyle \lim_{x\to 1}\frac{x^j-1}{x-1}=\lim_{x\to 1}(x^{j-1}+x^{j-2}+…+1)=j$$.

So our limit equals

$$\displaystyle \lim_{x\to 1}\frac{x+x^2+…+x^n-n}{x-1}=1+2+3+…+n=\frac{1}{2}n(n+1)$$,

where we have used the well-known equation about the sum of the first $$n$$ natural numbers.

• #### Result

The limit equals $$\frac{n(n+1)}2$$.