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 \(\frac{n(n+1)}2\).

Difficulty level: Easy task (using definitions and simple reasoning)
Routine calculation training
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