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Let $a_1=1$ , $a_{n+1}=a_n+(-1)^n \cdot 2^{-n}$ , $b_n=\frac{2 a_{n+1}-a_n}{a_n}$

(1) $\{\ {a_n\}}$ converges to $0$ and $\{\ {b_n\}}$ is a cauchy sequence .

(2) $\{\ {a_n\}}$ converges to non-zero number and $\{\ {b_n\}}$ is a cauchy sequence .

(3) $\{\ {a_n\}}$ converges to $0$ and $\{\ {b_n\}}$ is not a cauchy sequence .

(4) $\{\ {a_n\}}$ converges to non-zero number and $\{\ {b_n\}}$ is not a cauchy sequence .

Trial: Here $$\begin{align} a_1 &=1\\ a_2 &=a_1 -\frac{1}{2} =1 -\frac{1}{2} \\ a_3 &= 1 -\frac{1}{2} + \frac{1}{2^2} \\ \vdots \\ a_n &= 1 -\frac{1}{2} + \frac{1}{2^2} -\cdots +(-1)^{n-1} \frac{1}{2^{n-1}}\end{align}$$ $$\lim_{n \to \infty}a_n=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$$ Here I conclude $\{\ {a_n\}}$ converges to non-zero number. Am I right? I know the definition of cauchy sequence but here I am stuck to check. Please help.

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yes, you are right. Since $a_n$ is convergent sequence, so $b_n$ also convergent sequence. And if $x_n$ is convergent sequence, $x_n$ is cauchy. – Hanul Jeon Jan 21 '13 at 6:51
up vote 3 down vote accepted

We have $b_n=\frac{2 a_{n+1}-a_n}{a_n}=2\frac{a_{n+1}}{a_n}-1$. For very large values of $n$, since $a_n\to2/3$ we have $a_{n+1}\sim a_n$. So $b_n\to 2-1=1$ so it is Cauchy as well.

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Nice observations, +1 – amWhy Feb 11 '13 at 0:07

Your argument for the first part is correct $\lim\limits_{n\to\infty}a_n=\frac23$.

For the second part, $b_n=2\frac{a_{n+1}}{a_n}-1$. Since $a_n$ converges to a non-zero limit, $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$. Therefore, $\lim\limits_{n\to\infty}b_n=2\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}-1=2\cdot1-1=1$. Every convergent sequence is Cauchy, so $b_n$ is Cauchy.

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