I need help with this problem:
Let $f$ be a continous function over an interval that contains $0$. Let $a_n = f(\frac1 n)$ (for n large enough).
- Show that if $\sum_{n=1}^\infty \ a_n$ converges, then $f(0)=0$.
- Show that if $f'(0)$ exists and $\sum_{n=1}^\infty \ a_n$ converges, then $f'(0)=0$.
- Show that if $f''(0)$ exists and $f(0)=f'(0)=0$, then $\sum_{n=1}^\infty \ a_n$ converges.
- Suppose that $\sum_{n=1}^\infty \ a_n$ converges. Should $f'(0)$ exist?
- Suppose that $f(0)=f'(0)=0$. Should $\sum_{n=1}^\infty \ a_n$ converge?
I don't know how to solve this. I solved the first one, saying that if $\sum_{n=1}^\infty \ a_n$ converges that tells me that $\lim_{n\to\infty} a_n =0$. Beacuse of that, I can say that $\lim_{n\to\infty} f(\frac 1 n) = k$, since $f$ is continous on $0$, that means that $\lim_{x\to 0} f(\frac 1 n) = f(0)$ thus $f(0) = \lim_{x\to 0} f(x) = \lim_{n\to 0} f(\frac 1 n) = 0$.
How can I solve this? Please explain it to me.
Part (1) was answered in the comments.
For (2), suppose that $f'(0)=t> 0$. Then $$t=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{n\to\infty}\frac{f(1/n)-f(0)}{1/n}=\lim_{n\to \infty}nf(1/n)=\lim_{n\to\infty} na_n$$
Because $\lim_{n\to\infty} na_n=t$, there must be some $N$ such that for all $n>N$, $|na_n-t|<\frac{t}{2}$. Hence, for such $n$, $-\frac{t}{2}<na_n-t<\frac{t}{2}$. Adding $t$ throughout, we get $\frac{t}{2}<na_n$.
Hence, there is some $N$, such that for all $n>N$, we must have $na_n>\frac{t}{2}$, and hence $a_n>\frac{1}{n}\frac{t}{2}$. This gives $$\sum_{n>N} a_n \ge \sum_{n>N}\frac{1}{n}\frac{t}{2}=\frac{t}{2}\sum_{n>N} \frac{1}{n}$$ But the harmonic series diverges, so the original series diverges. Contradiction.
Note: if $t<0$, the proof is similar, except now $\sum a_n$ goes to $-\infty$.
For (3)-(5), you really should ask these as separate questions.