Show that as $N \to \infty$, $\sum_{i=N}^\infty{1} \to 0?$ How do I show that as $N \to \infty$, that
$$\sum_{i=N}^\infty{1} \to 0?$$
Don't know how to even start. Thanks..
Apparently this is wrong. But my teacher said that if $P_n f = \sum_{j=0}^n(f,w_j)w_j$, where $w_j$ is orthonormal basis of $L^2$, then $|P_n f- f|_{L^2} \to 0$. How can that be then? Because I thought
$$|P_nf - f| = |\sum_{j=0}^n(f,w_j)w_j - \sum_{j=0}^\infty(f,w_j)w_j| = |\sum_{j={n+1}}^\infty(f,w_j)w_j| \leq \sum_{j={n+1}}^\infty|f|$$
where the last equality is by Cauchy Schwarz.
 A: The limit does not exist. 
Consider the summation $\sum_{j=N}^{M-1} 1 = M- N$ for $M > N$. Roughly, you are after $\lim_{N \rightarrow \infty} \lim_{M \rightarrow \infty} M - N$. 
This limit does not exist. You can reason this out formally by expanding out the meaning of a limit diverging to $\infty$.  What I have below is most likely overkill for this problem.
Roughly, $\lim_{x \rightarrow \infty} f(x) = \infty$ means that $\forall K\ \exists x_0\ \forall x \geq x_0, f(x) > K$.
In this case, you want to show that 
$ (\forall\ K \geq 0)\ (\exists N_0 \geq 0)\ (\forall N \geq N_0)\ (\exists M_0 \geq 0)\ (\forall M \geq M_0),\ M - N > K$.
The statement above follows by setting $N_0$ to be some fixed number and $M_0 \geq N + K+1$.
A: $$\lim\limits_{N\to\infty}\left(\sum_{i=N}^{+\infty}1\right)=+\infty$$
A: When you say
$$\sum_{i=N}^\infty s_i$$
What you really have is a nice shorthand for:
$$lim_{M\to\infty} \sum_{i=N}^Ms_i$$
You can't just extend the notion of a sum, which is a combination of two elements in a group to a third, to a sum of an infinite number without some sort of limiting process. So what you're looking for is a limit that doesn't exist.
