# limit of $a_n(x)=\frac{\sum_{k=0}^{t_n}x^k}{\sum_{k=n+1}^{t_n}x^k}$

Let $t>1$ and $[t_n]$ defines the integer part of the product $t.n$. I need to evaluate the sequence $$a_n(x)=\frac{\sum_{k=0}^{[t_n]}x^k}{\sum_{k=n+1}^{[t_n]}x^k}$$ is convergent or bounded as $n\to\infty.$ Or for which values of $x$ this sequence is convergent or bounded ?

The easiest way to look at this problem is to convert the numerator and denominator from the presented sums to ratios using:

$$\Sigma_{k=0}^n x^k = {{1 - x^{n+1}} \over {1 - x}}$$

Using some standard manipulations, use can then replace your sequence with:

$$a_n(x) = {{1-x^{[t_n]+1}} \over {x^{n+1}(1-x^{[t_n]-n})}} = {{x^{[t_n]+1}-1} \over {x^{n+1}(x^{[t_n]-n}-1)}}$$

There are two cases: $|x|<1$ and $|x|>1$.

For $|x|<1$, the dominant part of each ratio is 1 and therefore your ratio becomes $a_n(x) \approx {1 \over x^{n+1}}$ which clearly diverges.

For $|x|>1$, the dominant part of each ratio is $x^m$ portion and therefore your ratio becomes: $$a_n(x) \approx { x^{[t_n]+1} \over x^{n+1}(x^{[t_n]-n)} } = {x^{[t_n]+1} \over x^{[t_n]+1} } = 1$$ Therefore for this range your sequence will converge to 1.