Recent content by fmam3

If you know liminf, limsup, then this is easy. Since a_n \leq b_n, or 0 \leq b_n - a_n. Then it follows that 0 \leq \liminf (b_n - a_n) \leq \limsup (b_n - a_n). But since \lim a_n = A and \lim b_n = B, it follows that \lim (b_n - a_n) = B - A, which implies the limit inferior and limit...

No prob :)

Note that WLOG, we can assume that A \ne 0 since if A = 0, then the result is trivial. In fact, we can even further say that we can assume, WLOG that A > 0 (note the strict inequality). Then, if we have \sqrt{a_n} + \sqrt{A} > A , then does that not imply \frac{1}{\sqrt{a_n} + \sqrt{A}} <...

Get the | |a| - |b| | \leq |a - b| right first. The second part just follows by the Squeeze Theorem, or more simply, just "shove in the epsilons" if you know what I mean.

Suppose \lim a_n = A and you want to show \lim \sqrt{a_n} = \sqrt{A}. So, right now you need a bound for the denominator. Assuming that we are working with the reals (i.e. no complex numbers for this problem), then see that for the square root to be defined, we must have that a_n, A \geq 0...

Actually, an even easier argument could be made as follows. Since \liminf |a_n| = 0 and since the liminf is the infimum of the set of subsequential limits, there exists a subsequence (|a_{n_k}|) such that \lim_{k \to \infty} |a_{n_k}| = 0. This means that \forall \varepsilon > 0, \exists N > 0...

To clear up the OP's problem statement: Let (a_n) be a sequence of reals such that \liminf |a_n| = 0. Prove that there is a subsequence (a_{n_k}) such that the series \sum_{k=1}^{\infty} a_{n_k} converges. Let \forall \varepsilon > 0. Since \liminf |a_n| = 0, there \exists N > 0 such that...

Actually, liminf and limsup (i.e. limit inferior and limit superior) are precisely defined as follows. Let (s_n) be a sequence of real numbers. Then define u_N = \inf\{s_n : n > N\}. Then \liminf s_n = \lim_{N \to +\infty}u_N. Limit superior is defined analogously. There are many ways of...

I don't think its entirely correct to say that the sequence (a_n) converges to zero. Note that it is only given that \liminf |a_n| = 0 and we have \lim |a_n|= 0 if and only if \liminf |a_n| = \limsup |a_n|.

Thanks again to both Billy Bob and Office_Shredder! You guys were great help :)

Ah... yes perfect! Yes, I should have caught that error myself (i.e. x \in (0,1) part). Then we have that h(x) < h(1) = 0 for \exists x \in (0,1). And since h(0) = g(0) - 0 > 0, it follows that we have h(x) < 0 < h(0) for \exists x \in (0,1). Hence, since h is clearly continuous on [0,1], then...

Thanks for the reply again. I mean, geometrically, I see exactly what you mean. But since it is necessary to prove this rigorously, I'm sure there must be a way of applying the Mean Value Theorem (and I'm dead sure it must be used in the converse case) to solve it. *sigh*... But thanks for...

I'm actually wondering --- can my proof still be repaired? So if we start of by assuming that \forall d \in (0,1), g(d) = d, then we must have the two cases: (1) \forall d \in (0,1), g(d) \geq d; and (2) \forall d \in (0,1), g(d) \leq d Now, if I were two replace in my above proof g(d) > d...

Yes... you're right. I screwed up on the negation. The negation of the statement "\exists d \in (0,1) s.t. g(d) = d" should be \forall d \in (0,1), g(d) = d. Actually, I showed that g'(c_1) < 1 in the above. But this doesn't matter, since my negated statement is wrong anyways. I'll have...

Thanks Billy Bob & Office_Shredder for the tips above! I think I have figured out a solution. Here it is :) For the converse, suppose we have g'(1) > 1 and we want to show g(d) = d for some d \in (0,1). For contradiction, suppose we have g(d) \ne d for \forall d \in (0,1). Suppose...