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"Prove that $ \lim_{n \to \infty} \, \, \left(\frac{1}{1+a_n} \right) = \frac{1}{2}$ if $\lim_{n \to \infty} a_n = 1$."

I understand the algebra, but when I get to this step:

$ |1-a_n|\left| \frac{1}{2(1+a_n)} \right| < \epsilon $

I have no idea what to do. Am I allowed to just divide the right-sided product to the epsilon side? Likewise, I don't think I can bound the right-sided product, since if $-1 < a_n < -0.5$ it explodes. I am so close to getting what I want, but don't know how to get there. Any help please?

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  • $\begingroup$ Remember you need to make use of the fact that $a_n \rightarrow 1$. Since $a_n \rightarrow 1$, given $\epsilon > 0$, there exists $N$ such that $\forall n > N$, $\left|a_n - 1 \right| < \epsilon$. Make use of this and also bound $\left| \frac1{2(1+a_n)} \right|$ by $1$ (Why?). $\endgroup$ – user17762 Mar 14 '12 at 5:32
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Ok, here is what you do, you can choose $N$ to be large enough so that $a_n$ is say $0.5$-close to $1$ for all $n\geq N$. Then after having chosen this $N$, you can bound it as follows: $$|1-a_n||\frac{1}{2(1+a_n)}|<|1-a_n|\frac{1}{2}$$ since you are decreasing the numerator you are overall increasing the fraction. Lastly, you can choose $N'$ to be large enough (make it larger that $N$) so that $|1-a_n|<2\epsilon$, and the result follows.

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  • $\begingroup$ It is implicit that if $a_n$ is $0.5$-close to $1$, then $a_n$ is strictly positive and when we change denominator $1+a_n$ for $1$ is a decrease on the denominator. Im just saying it in case it wasnt completely clear. $\endgroup$ – Daniel Montealegre Mar 14 '12 at 5:34
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Let $\epsilon_1=\min(\epsilon, 1/2)$.

There is an $N$ such that if $n >N$, then $|a_n-1|<\epsilon_1$.

But then if $n>N$, $$|1-a_n|\left|\frac{1}{2(1+a_n)}\right|<\epsilon.$$

Remark: You saw the issue: if we allow $a_n$ to roam to near $-1$, then we cannot control $\left|\frac{1}{1+a_n}-\frac{1}{2}\right|$. The fix is straightforward. One just picks $N$ large enough that $a_n$ is near $1$. Some version of this idea is often needed.

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