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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!

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  • $\begingroup$ Your argument is correct. The limit is indeed $\frac{1}{2}k(k+1)$ $\endgroup$ – Crostul Jan 15 at 16:05
  • $\begingroup$ Not to dampen your spirit, but perhaps a proof using L'Hopital would be shorter? $\endgroup$ – Keen-ameteur Jan 15 at 16:11
  • $\begingroup$ @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. $\endgroup$ – roman Jan 15 at 16:13
  • $\begingroup$ Okay, then nevermind. $\endgroup$ – Keen-ameteur Jan 15 at 16:14
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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.

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  • $\begingroup$ That one is a nice approach $\endgroup$ – roman Jan 15 at 16:16
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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*}

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

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