Limit of seriess

Question

Be $x_n = ac+\left(a+ab\right)c^2+...+\left(a+ab+...+ab^n\right)c^{n+1}$.

If $|c|<1$ , $b\neq 1$ and $|bc|<1$ then $\lim \limits_{n\to \infty }x_n=?$

It's a multiple choice question and i don't even know how to treat it. It would be very helpful if you could treat some of the possible answers.

A) $(x_n)$ does not converge

B) $\lim \limits_{n\to \infty }x_n=0$

C) $\lim \limits_{n\to \infty }x_n=\frac{\left(a+bc\right)}{\left(1-ab\right)c}$

D) $\lim \limits_{n\to \infty }x_n=1$

E) $\lim \limits_{n\to \infty }x_n=\frac{ac}{\left(1-bc\right)\left(1-c\right)}$

Answer 1

If $ a = 0$ then $ x_n = 0 $ so the limit is $ 0 $. If $ a \neq 0 $, dividing by $ a $ we have $$ x_n / a = c + (1 + b)c + ... + (1 + b + b^2 + ... + b^n )c^{n + 1} $$ Now we observe that $$ 1 + b + ... + b^n = \frac{b^{n + 1} - 1}{b - 1} \ \ \mbox{ for each } n \in \mathbb{N} $$ because the expression is the sum of the $ n $ first terms of a geometric progression. Then the equality became as $$ x_n / a = \frac{b - 1}{b - 1} c + \frac{b^2 - 1}{b - 1} c + ... + \frac{b^{n + 1} - 1}{b - 1} c^{n + 1} = $$ $$ \frac{1}{b - 1} \cdot \left( (b - 1)c + (b^2 - 1)c^2 + ... + (b^{n + 1} - 1)c^{n + 1} \right) $$ The expression can be transform in a sum of two geometric progression $$ (b - 1)c + (b^2 - 1)c^2 + ... + (b^{n + 1} - 1)c^{n + 1} = (bc - c) + ( (bc)^2 - c^2) + ... + ( (bc)^{n + 1} - c^{n + 1} ) = $$ $$ (bc + (bc)^2 + ... + (bc)^{n + 1}) - (c + c^2 + ... + c^{n + 1} ) $$ Now we use again that the two sums are geometric progressions to obtain $$ bc + (bc)^2 + ... + (bc)^{n + 1} = \frac{(bc)^{n + 2} - 1}{bc - 1} - 1 \ \ \mbox{ and } \ \ c + c^2 + ... + c^{n + 1} = \frac{c^{n + 2} - 1}{c - 1} - 1 $$ Then we have $$ x_n /a = \frac{1}{b - 1} \cdot \left( \frac{(bc)^{n + 2} - 1}{bc - 1} - 1 - \left( \frac{c^{n + 2} - 1}{c - 1} - 1 \right) \right) = \frac{1}{b - 1} \cdot \left( \frac{(bc)^{n + 2} - 1}{bc - 1} - \frac{c^{n + 2} - 1}{c - 1} \right) $$ We know that $ \mid bc \mid < 1 $ and $ \mid c \mid < 1 $, so the expressions $ (bc)^{n + 1} \rightarrow 0 $ if $ n \rightarrow \infty $ and $ (c)^{n + 1} \rightarrow 0 $ if $ n \rightarrow \infty $. Then we can take $ \lim_n $ on both sides to obtain $$ \lim_n x_n / a = \frac{1}{b - 1} \cdot \lim_n \left( \frac{(bc)^{n + 2} - 1}{bc - 1} - \frac{c^{n + 2} - 1}{c - 1} \right) = \frac{1}{b - 1} \cdot \left( -\frac{1}{bc - 1} + \frac{1}{c - 1} \right) $$ So, $$ \lim_n x_n = \frac{a}{b - 1} \left( \frac{-(c - 1) + (bc - 1)}{ (c - 1)(bc - 1) } \right) = \frac{a}{b - 1} \cdot \frac{ c(b - 1) }{(bc - 1)(c - 1)} = \frac{ac}{(bc - 1)(c - 1)} $$ In the last step we can cancel $ b - 1 $ because $ b - 1\neq 0 $. This is the E) option.

Answer 2

Hint: Write the coefficient of $c^k$ in closed form. Each such coefficient is a finite geometric series with $k+1$ terms involving $a$ and $b$.

Answer 3

$$ \begin{align} \lim_{n\to\infty}x_n &=\sum_{k=0}^\infty a\,\frac{b^{k+1}-1}{b-1}\,c^{k+1}\\ &=\frac{a}{b-1}\left[\sum_{k=0}^\infty (bc)^{k+1}-\sum_{k=0}^\infty c^{k+1}\right]\\ &=\frac{a}{b-1}\left[\frac{bc}{1-bc}-\frac{c}{1-c}\right]\\[6pt] &=\frac{ac}{(1-bc)(1-c)} \end{align} $$

Answer 4

The general form of your sequence is given by

$$x_n = \sum_{k=0}^{n}\left(ac^{k+1}\sum_{i=0}^{k}b^i\right)$$

which, using the partial sum formula for the geometric series, simplifies to

\begin{align}x_n &= a\sum_{k=0}^{n}\frac{c^{k+1}}{1-b}\left(1-b^{k+1}\right)\\ &=\frac{a}{1-b}\sum_{k=0}^{n}c^{k+1}-c^{k+1}b^{k+1}\\ &=\frac{a}{1-b}\left(\sum_{m=1}^{n+1}c^k - \sum_{m=1}^{n+1}(cb)^k\right)\\ &=\frac{a}{1-b}\left(-1+\sum_{m=0}^{n+1}c^k + 1 -\sum_{m=0}^{n+1}(cb)^k\right)\\ &=\frac{a}{1-b}\left(\sum_{m=0}^{n+1}c^k - \sum_{m=0}^{n+1}(cb)^k\right) \end{align}

Since $|c|, |bc|<1$, both sequences of sums converge and one can use the formula for the geometric series again to calculate their value.

Thus

\begin{align} \lim\limits_{n\rightarrow\infty}x_n&=\frac{ac}{1-b}\left(\frac{1}{1-c}-\frac{1}{1-bc}\right)\\ &=\frac{a}{1-b}\frac{1-bc-1+c}{(1-c)(1-bc)}\\ &=\frac{a}{1-b}\frac{c(1-b)}{(1-c)(1-bc)}\\ \end{align}

Finally, cancelling out a factor $1-b$, we arrive at

$$\lim\limits_{n\rightarrow\infty}x_n=\frac{ac}{(1-bc)(1-c)}$$

therefore answer $E)$ is the correct one.