Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  tetori Jan 21 '13 at 6:51

2 Answers 2

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.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.