Limit of the sum of a series around 1
Task number: 2977
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 \(\frac{n(n+1)}2\).
