The problem here is, that you use the fact that if p|(n+M) \text{ and } p|n \Rightarrow p|M. But this doesn't cover all the possibilities, namely if p\not|n \text{ and } p\not|M \text{ but } p|(n+M), as micromass pointed out.
For any k\geq 2 choose n to be the product of all prime numbers \leq k. Then \{p\in\mathbb{P} : p < n, p \not |n \}=\emptyset. So M need not exist, if I'm not mistaken.
Yeah, that's the part that caught me as well. I remember seing your function somewhere also, it seems to be the most straight forward. The function sequence I had in mind was
f_n(x) =
\begin{cases}
n\cdot x & \text{if} \; \; 0 \leq x < \frac{1}{n} \\
1 & \text{if} \; \; \frac{1}{n} \leq x...
Under the orthogonality section in the wikipedia article on Legendre polynomials, you find the identity
\displaystyle \frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P(x)\right] = -\lambda P(x)
where the eigenvalue \lambda corresponds to n(n+1).
I suppose P(x)=P_n(x) for any n, but I'm not sure though...
Mayby I'm confused with the notations but I thought that \displaystyle d_{\infty}(f,g)=\sup_{x\in[a,b]}|f(x)-g(x)|<\epsilon meant uniform convergence, since doesn't this mean that
\forall x\in[a,b]:\ |f(x)-g(x)|<\epsilon?
For the last problem, try to find a sequence of continuous functions that converges to a non-continuous function. I have one in mind, but haven't thought of the details yet. Consider piecewise functions...
edit: Sorry, the path I suggested might not lead you anywhere. It seems that if a...
What properties do you know about Legendre Polynomials? If you can use the orthogonal properties that are listed in the article on Legendre polynomials in wikipedia, then integration by parts should do the trick.
I suppose you would prove it by the same reasoning as you'd prove that the sequence \{n\}_{n\in \mathbb{N}} diverges, reductio ad absurdum using the definition of convergence to get a contradiction.
If it is calculated like I suspect it is, then you will get a satisfying answer.
So first we want to find out what f_3^n is when n\geq 3:
f_3^n = \displaystyle \frac{1}{2}\cdot1\cdot\left(\prod_{k=3}^{n-1}\left(\frac{k}{k+1}\right)^a\right)\cdot\left(1-\left(\frac{n}{n+1}\right)^a\right) =...