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Thread: convergence

  1. #1
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    convergence

    If $\displaystyle a_n $ is a sequence such that $\displaystyle \frac{a_{n}-1}{a_{n}+1} \to 0 $, then does $\displaystyle \lim_{n \to \infty} a_{n} $ exist? So in other words, does $\displaystyle a_n \to L $?

    Assume that $\displaystyle s_n = \frac{a_{n}-1}{a_{n}+1} $. Let $\displaystyle \varepsilon > 0 $. Then there exists an $\displaystyle N \in \mathbb{N} $, such that whenever $\displaystyle n \geq N $, then $\displaystyle |s_{n}| < \varepsilon $.

    Now the problem is to express $\displaystyle a_{n} $ in terms of $\displaystyle s_{n} $ and use the definition of convergence to see if it converges or not?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by particlejohn View Post
    If $\displaystyle a_n $ is a sequence such that $\displaystyle \frac{a_{n}-1}{a_{n}+1} \to 0 $, then does $\displaystyle \lim_{n \to \infty} a_{n} $ exist? So in other words, does $\displaystyle a_n \to L $?

    Assume that $\displaystyle s_n = \frac{a_{n}-1}{a_{n}+1} $. Let $\displaystyle \varepsilon > 0 $. Then there exists an $\displaystyle N \in \mathbb{N} $, such that whenever $\displaystyle n \geq N $, then $\displaystyle |s_{n}| < \varepsilon $.

    Now the problem is to express $\displaystyle a_{n} $ in terms of $\displaystyle s_{n} $ and use the definition of convergence to see if it converges or not?
    If you have to use the definition of convergence, that's the idea. To help you, $\displaystyle |s_n|<\varepsilon$ can be written $\displaystyle -\varepsilon < 1-\frac{2}{a_n+1} < \varepsilon$.

    If you don't have to use the definition of convergence, suppose that $\displaystyle (a_n)$ diverges and see what happens to $\displaystyle s_n=1-\frac{2}{a_n+1}$ which is supposed to tend to 0 as $\displaystyle n\to\infty$.
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