<s>[INST] Sum limit. Please tell if correct. I just solved this limit, and the result provided by the book is different.
$$ \lim_{x\to 1} \frac{x+x^2+x^3 + ... + x^n - n}{x-1} $$
I turned this into:
$$ \lim_{x\to 1} \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \frac{x^3-1}{x-1} + ...  + \frac{x^n-1}{x-1} $$
$$ \lim_{x\to 1} 1 + x + (x + x^2) + (x + x^2 + x^3) +...+(x + x^2 + ... +         x^{n-1})$$
$$ 1 + 1 + 2 +3 +...+ n-1 $$
My result: $ \frac{n(n-1)}{2}+1 $ 
Books result: $ \frac{n(n-1)}{2} $
Have i missed something?
 [/INST] Your method of solving is correct and nice, but there are some mistakes. First
$$\frac{x^k-1}{x-1}=x^{k-1}+x^{k-2}+\ldots+x+1$$
and thus
$$\frac{x+x^2+\ldots+x^n-n}{x-1}=\sum_{k=1}^n\frac{x^k-1}{x-1}=$$
$$=\sum_{k=1}^n\left(1+x+\ldots+x^{k-1}\right)=1+2+\ldots+n=\frac{n(n+1)}2$$
One can also do the following, though perhaps you haven't yet studied this: define
$$f(x)=x+x^2+\ldots+x^n\;,\;\;n\in\Bbb N\;\;\text{a constant}$$
Then your limit is this functions' derivative at $\;x=1\;$ :
$$\lim_{x\to1}\frac{x+x^2+\ldots+x^n-n}{x-1}=\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=f'(1)$$
and since $\;f'(x)=1+2x+3x^2+\ldots+nx^{n-1}\;$ , we get that
$$f'(1)=1+2+3+\ldots+n=\frac{n(n+1)}2$$
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