I'm working on a homework assignment right now and I've hit a snag. We've been given a true statement and a fake proof:
The professor wants us to rewrite the proof and say what the limit method actually proves, but I'm confused about why this proof is actually invalid?
Obviously this proof is easy enough to solve on it's own, since it evaluates to cn$^k$ $\in$ cn$^k$$^+$$^1$. But I don't quite understand why this limit method is failing. The limit evaluates to 0, and I think all the calculus is correct.
Any help would be very appreciated!
(1). It is not generally true that $\lim_{n\to \infty} \sum_{i=1}^n A(i,n)=\sum_{i=1}^{\infty}\lim_{n\to \infty}A(i,n).$ This is the logical error in the "fake proof".
A simple example is to let $A(n,n)=1$ and to let $A(i,n)=0$ when $i\ne n.$ Then $\sum_{i=1}^nA(i,n)=1$ for each $n$ but $\lim_{n\to \infty}A(i,n)=0$ for each $i.$
(2). For $k\geq 0$ and $m\in \Bbb N$ we have $$\int_{m-1}^mx^kdx \leq\int_{m-1}^mm^kdx =$$ $$=m^k=$$ $$= \int_m^{m+1}m^kdx\leq \int_m^{m+1}x^kdx .$$ Adding from $m=1$ to $m=n,$ we have $$\frac {n^{k+1}}{k+1}=\int_0^nx^kdx\leq$$ $$\leq \sum_{m=1}^n m^k\leq$$ $$\leq \int_1^{n+1} x^kdx=\frac {(n+1)^{k+1}-1}{k+1}.$$ From this we readily obtain $\lim_{n\to \infty}\frac {f(n)}{g(n)}=\frac {1}{k+1}.$