# Find the limit $\lim_{n\to\infty}\frac{x_n + x_n^2 + \cdots + x_n^k - k}{x_n - 1}$ 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!

• Your argument is correct. The limit is indeed $\frac{1}{2}k(k+1)$ – Crostul Jan 15 '19 at 16:05
• Not to dampen your spirit, but perhaps a proof using L'Hopital would be shorter? – Keen-ameteur Jan 15 '19 at 16:11
• @Keen-ameteur Unfortunately I'm not supposed to use derivatives or even Stolz-Cesaro since none of them has been yet defined at where i am now in the book. – roman Jan 15 '19 at 16:13
• Okay, then nevermind. – Keen-ameteur Jan 15 '19 at 16:14

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.

• That one is a nice approach – roman Jan 15 '19 at 16:16

Let $$y_n=x_n-1$$ and then $$\lim_{n\to\infty}y_n=1$$. Note $$\begin{eqnarray*} &&\frac{x_n + x_n^2 + \cdots + x_n^k - k}{x_n-1}\\ &=&\frac{(y_n+1)^{k+1}-(k+1)(y_n+1)+k}{y_n^2}\\ &=&\frac{\sum_{i=0}^{k+1}\binom{k+1}{i}y_n^i-(k+1)y_n-1}{y_n^2}\\ &=&\frac{\sum_{i=2}^{k+1}\binom{k+1}{i}y_n^i-(k+1)y_n}{y_n^2}\\ &=&\sum_{i=2}^{k+1}\binom{k+1}{i}y_n^{i-2}\\ \end{eqnarray*}$$ and hence $$\begin{eqnarray*} &&\lim_{n\to\infty} \frac{x_n + x_n^2 + \cdots + x_n^k - k}{x_n - 1}\\ &=&\lim_{n\to\infty}\sum_{i=2}^{k+1}\binom{k+1}{i}y_n^{i-2}\\ &=&\binom{k+1}{2}=\frac12k(k+1). \end{eqnarray*}$$

This could be realized as a derivative. Let us define $$f(x)=x^1+x^2+\cdots+x^k\,.$$ Then $$\lim_{x\rightarrow 1}\frac{f(x)-f(1)}{x-1}=f'(1)=1+2\cdot 1^1+\cdots+k\cdot 1^{k-1}=\frac{k(k+1)}{2}\,.$$