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here's my daily problem:

1) $b_n=|a_n| + 1 - \sqrt {a_n^2+1}$. I have to prove that, if $b_n$ converges to 0, then $a_n$ converges to 0 too. Here's how I have done, could someone please check if this is correct? I'm always afraid to square both sides.

$ \begin{align*} 0&=|a_n| + 1 - \sqrt {a_n^2+1}\\ & -|a_n| = 1 - \sqrt {a_n^2+1}\\ & a_n^2 = 1 - 2 *\sqrt {a_n^2+1} + a_n^2 + 1\\ & 2 = 2 *\sqrt {a_n^2+1}\\ & 1 = \sqrt {a_n^2+1}\\ & 1 = {a_n^2+1}\\ & a_n^2 = 0 \Rightarrow a_n=0\\ \end{align*}$

2) $b_n = \frac{|a_n|}{1+|a_{n+2}|}$ I have to prove the following statement is false with an example: "If $b_n$ converges to 0, then $a_n$ too." I'm pretty lost here, any directions are welcome! I thought that would only converge to 0, if $a_n=0$. Maybe if $a_n >>> a_{n+2}$?

Thanks in advance! :)

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While your algebra in 1) looks ok, you cannot assume $b_n=0$ only that it is converging to 0. Unravel the $\epsilon-\delta$-definition of limit and then use your algebra to make a conclusion about $a_n$. –  Bill Cook Nov 9 '11 at 19:29
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2 Answers

up vote 3 down vote accepted

1) One has $b_n=\dfrac{2|a_n|}{|a_n|+1+\sqrt{a_n^2+1}}\geqslant\dfrac{|a_n|}{1+|a_n|}\geqslant0$.

Assume that $b_n\to0$. Then the inequality above shows that $\dfrac{|a_n|}{1+|a_n|}\to0$. If $\dfrac{|a_n|}{1+|a_n|}\leqslant\varepsilon$ with $\varepsilon\leqslant\frac12$, then $|a_n|\leqslant2\varepsilon$. Hence $|a_n|\to0$, which is equivalent to $a_n\to0$.

2) Try $a_n=2^{n^2}$.

Edit As regards the example suggested above for 2), it seems that a cul-de-sac was reached in the comments (see below, I cannot do any better to explain why what you say is false than what I already did), so let me try another example: let $a_n=n!$.

Then $1+|a_{n+2}|\gt |a_{n+2}|=(n+2)(n+1)|a_n|$ hence $0\lt b_n\lt \dfrac{|a_n|}{|a_{n+2}|}=\dfrac1{(n+1)(n+2)}\to0$. One sees that $b_n\to0$ although $a_n\to+\infty$.

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Thanks for the switf reply didier! The result I got with your example for 2) was $\frac{1}{2^{2*n+4}}$, which converges to 0. Can you please confirm I got it right? Thanks! –  Clash Nov 9 '11 at 20:43
    
Result of what, in relation with 2)? In any case if the $a_n$ are powers of $2$, the ratio $|a_n|/(1+|a_{n+2}|)$ is not a power of $2$. –  Did Nov 10 '11 at 11:13
    
I did $a_n=2^{n^2}$ at $|a_n|/(1+|a_{n+2}|)$ and got $\frac{1}{2^{2*n+4}}$, which converges to $0$. Is this false? –  Clash Nov 10 '11 at 13:52
    
If $a_n$ and $a_{n+2}$ are (positive) powers of $2$, $|a_n|$ is a power of $2$ and $1+|a_{n+2}|$ is an odd integer hence the fraction $|a_n|/(1+|a_{n+2}|)$ is irreducible, in particular its value is not a (negative) power of $2$. But if your goal is to show that $b_n\to0$, since $b_n$ is nonnegative, it is enough to find some simpler $c_n$ such that $b_n\leqslant c_n$ and $c_n\to0$. Any idea for what $c_n$ might be? –  Did Nov 10 '11 at 16:18
    
Any progress on this $c_n$ stuff? –  Did Nov 11 '11 at 18:25
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You seem to have a fundamental misconception regarding the difference between the limit of a sequence and an element of a sequence.

When we say $b_n$ converges to $0$, it does not mean $b_n = 0$ for all $n$. For instance $b_n = \frac{1}{n}$ is convergent to $0$, but there is no natural number $n$ for which $b_n = 0$.

In i) What you tried is ok (though the misconception above shows in the way you have written it), but are making some assumptions which need to be proved.

If we were to rewrite what you wrote, it would be something like,

If $a_n$ was convergent to $L$, then we would have that

$ 0 = |L| + 1 - \sqrt{L^2 + 1}$

and then the algebra shows that $L = 0$.

Using $a_n$ instead of $L$ makes what you wrote nonsensical.

Also, can you tell what assumption is being made here and needs justification?

For ii) Try constructing a sequence such that $\frac{a_{n+2}}{a_n} \to \infty$.

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