calculus - Find the limit $lim_{ntoinfty}frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}$ given $x_n ne 1$ and $lim x_n = 1$

Let $x_n$ denote a sequence, $n\in\Bbb N$. Evaluate the limit:
$$
\lim_{n\to\infty} \frac{x_n + x_n^2 + \cdots + x_n^k - k}{x_n - 1},\ k\in\Bbb N

$$
given
$$
\lim_{n\to\infty} x_n = 1 \\
x_n \ne 1
$$

I'm interested in verifying the following results. Denote:
$$

y_n = \frac{x_n + x_n^2 + \cdots + x_n^k - k}{x_n - 1}
$$
After some trials long division seem to produce a pattern here. Not sure how to put it here in a fancy way, so I will post the result only. It appears that:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+\cdots + (k-1)x_n + k
$$

It is given that $\lim x_n = 1$ therefore we may use the following properties:
$$
\lim(a_n + b_n) = \lim a_n + \lim b_n\\

\lim(a_n \cdot b_n) = \lim a_n \cdot \lim b_n
$$

In particular:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+\cdots + (k-1)x_n + k\\
\lim_{n\to\infty}y_n =\lim_{n\to\infty}\left(x_n^{k-1} + 2x_n^{k-2}+\cdots + (k-1)x_n + k\right)
$$

Then since $\lim x_n = 1$ and by the sum of first $k$ integers:

$$
\lim_{n\to\infty} y_n = \frac{k(k+1)}{2}
$$

Could you please verify whether the reasoning above is correct or not and point to the mistakes in case of any? Thank you!

Answer

Rewrite the limit as

$$ \sum_{j=1}^k \lim_{n \to \infty} {x_n^j - 1 \over x_n - 1} $$

(note that this is a sum of finitely many terms).

Now we can do the division to get

$$ \sum_{j=1}^k \lim_{n \to \infty} (1 + x_n + x_n^2 + \cdots + x_n^{j-1}) $$

and the limit can be written as a sum of (again, finitely many!) limits. This gives

$$ \sum_{j=1}^k \sum_{i=0}^{j-1} \lim_{n \to \infty} x_n^i $$

Finally the innermost limit is, for each $n$ and $i$, equal to $(\lim_{n \to \infty} x_n)^i = 1^i = 1$ so this is just

$$ \sum_{j=1}^k \sum_{i=0}^{j-1} 1 = \sum_{j=1}^k j = {k(k+1) \over 2}$$

as desired.