Question: Show $b\_0$,.....,$b\_{n-1}$ are real numbers, then for $k$ sufficiently large,

$\frac{1}{2} < 1 + \frac{b\_{n-1}}{k} +$...$+\frac{b\_0}{k^n} < \frac{3}{2}$. 

Right now I am studying Tests of Converges of Series, so my partial solution tries this approach. please help me with my solution.

**solution:**

$\frac{1}{2} < 1 + \frac{b\_{n-1}}{k} +$...$+\frac{b\_0}{k^n} < \frac{3}{2}$. 

$\frac{k^n}{2} < k^n + b\_{n-1}k^{n-1} +$...$+b\_0< \frac{3k^n}{2}$. 

$\frac{k^n}{2} < k^n(1 + b\_{n-1}(\frac{1}{k}) +$...$+b\_0({\frac{1}{k^n}})) < \frac{3k^n}{2}$. 

Let $c\_k =1 + b\_{n-1}(\frac{1}{k}) +$...$+b\_0({\frac{1}{k^n}})$

$c\_k \leq \frac{3k^n/2}{k^n/2} = 3$

Let $a\_k = k^nc\_k \leq 3k^n$ for sufficiently large $k$.

Answer: 
Let $M$ be the maximum of the $|b\_i|$, and let $k\ge 1$. Then by the Triangle Inequality
$$\left|\frac{b\_{n-1}}{k}+\cdots +\frac{b\_0}{k^n}\right|\le \frac{nM}{k},$$ since $k^i\ge k$. 

Thus to make our inequalities hold, it is enough to choose $k\ge 1$ so large that $\frac{nM}{k}\lt \frac{1}{2}$.
Generalizing
André Nicolas's
answer:

As he showed,
$\left|\frac{b\_{n-1}}{k}+\cdots +\frac{b\_0}{k^n}\right|
\le \frac{nM}{k}
$.

For any $\epsilon > 0$,
if we choose
$k > \frac{nM}{\epsilon}
$,
then
$1-\epsilon
 < 1 + \frac{b\_{n-1}}{k} +$...$+\frac{b\_0}{k^n} 
<1+\epsilon
$,
so we can make the sum
as close to $1$
as we want.