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I have the following sequence (I've already put it in the $a_n - L$ form)
$$\left|\frac{(-1)^n n+1}{n + 2}-L\right|\ge\epsilon$$
I think that in order to prove that the sequence is divergent I need to prove that $\ge\epsilon$ instead of the usual $<\epsilon$
My problem is that I have L which is stuck there so Im not sure if Im supposed to make it go somehow or play along with it like its just a number which uglify my expression

The One
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2 Answers2

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$$ \frac{(-1)^n n+1}{n + 2} = \frac{(-1)^n}{1+2/n} + \frac{1}{n + 2} $$

The subsequence of odd indices converges to $-1$. The subsequence of even indices converges to $+1$.

For the rest, read this.

Community
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mvw
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Show that for all $\epsilon_{1,2}>0$ there are $N_{1,2}\in \mathbb N$ such that $ |a_{2n}-1|<\epsilon_1$ and $ |a_{2n+1}+1|<\epsilon_1$ for all $n>N_1,N_2$

Hence the sequence is divergent

That's a formal way but of course I started different as you... still okay for you?

Marm
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  • I think it is formal enough it just means that in order to prove divergence I need to prove that there are at least 2 differnt $\epsilon$ that apply to what you said? I cant find a clear definition of divergent sequence – The One Nov 15 '14 at 20:16
  • After you proved the existance of the two limit points, take $\epsilon_1=\epsilon_2=0.5$ and see whats happens – Marm Nov 15 '14 at 20:20
  • Divergent sequence in English just seems to mean "not convergent sequence". In German there is a distinction between "bestimmt divergent" (to $-\infty$, $+\infty$) and "unbestimmt divergent" (your case), which I can not find in English so far. – mvw Nov 15 '14 at 20:32
  • @mvw I can't even find the mathematical definition (besides the not converge defintion) – The One Nov 15 '14 at 20:41
  • A sequence with does not converge is called a divergent sequence :) So, just negate the definition of convergent sequence and you will get the solution together with my comment above. – Marm Nov 15 '14 at 20:44
  • I got $n> \frac{1}{\epsilon_1}$ and $n > \frac{3}{\epsilon_2}$ what am I supposed to see when the epsilons equal 0.5? – The One Nov 15 '14 at 21:56
  • If we choose \epsilon_{1,2}=0,5, then for a suitable $N\in \mathbb N$, we have $|a_{2n}-1|<0.5$ and $|a_{2n+1}+1|<0.5$, hence both are not limit points, but accumulation points. Try to understand by your own why it is like this..a sketch can help and recall the definiton of convergence. – Marm Nov 15 '14 at 23:30