Find the value of the following limit.
![\[ \lim_{x \to 1} \frac{ \sum_{k=1}^n x^k - n }{x-1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c8737f75741024eb8aa16e8cb531ba9a_l3.png "Rendered by QuickLaTeX.com")
First, we use the algebraic identity,
![\[ \sum_{k=1}^n x^k = \frac{1-x^{n+1}}{1-x} - 1\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9d8ef9a32e19c68c0b4f450c93548157_l3.png "Rendered by QuickLaTeX.com")
to obtain
![\begin{align*} \lim_{x \to 1} \frac{ \sum_{k=1}^n x^k - n }{x-1} &= \lim_{x \to 1} \frac{ \left( \frac{1-x^{n+1}}{1-x} \right) - 1 - n}{x-1} \\[9pt] &= \lim_{x \to 1} \frac{ 1 - x^{n+1} - (1+n)(1-x)}{(1-x)(x-1)} \\[9pt] &= \lim_{x \to 1} \frac{x^{n+1} - 1 + (1+n)(1-x)}{(x-1)^2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-eb0682f60fca7de51419d5db695dee2a_l3.png "Rendered by QuickLaTeX.com")
Then, we apply L’Hopital’s rule twice,
![\begin{align*} \lim_{x \to 1} \frac{\sum_{k=1}^ x^k - n}{x-1} &= \lim_{x \to 1} \frac{x^{n+1} - 1 + (1+n)(1-x)}{(x-1)^2} \\[9pt] &= \lim_{x \to 1} \frac{(n+1)x^n - (1+n)}{2(x-1)} \\[9pt] &= \lim_{x \to 1} \frac{n(n+1)x^{n-1}}{2} \\[9pt] &= \frac{n(n+1)}{2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f2c5ce5c2f203727e27764a525524f80_l3.png "Rendered by QuickLaTeX.com")
Find the value of the following limit.
First, we use the algebraic identity,
to obtain
Then, we apply L’Hopital’s rule twice,
Using Gauss sum to obtain n(n+1)/2 after applying LH’opital’s rule directly a little bit easier. This web site is really helpful for the ones self studying by the way