Three prove/disprove question in calculus 1) $f:(0,\infty$)$\to\Bbb R$ Continuously differentiable function which maintain 
$\lim_{x\to\infty} \ xf'(x)$=1 than $ \lim_{x\to\infty} \ f(x)$=$\infty$.
2) For every $n$,  $a_n>0$ and $\limsup_{n\to\infty} \ na_n = 1$ than $\sum_{n=1}^{\infty} a_n $ diverges.
3) For every $n$,  $a_n>0$  and $\sum_{n=1}^{\infty} a_n $  converges than  $\sum_{n=1}^{\infty} \max (a_n,a_{n+1}) $ converges. 
For the first question I think its correct since I tried every function imaginable I could think of, but I could not prove it.
For the second question with the condition $a_n>0$ I tried to use comparison theories but it did not help (assumed it is not correct but did not get to a contradiction)  so I figure it is false, but I could not find a counterexample with the condition $a_n>0$.
For the third I figure its false with a function similar to $\sin(n)$(not monotonic function in general) but again the condition $a_n>0$ sabotages it. 
Hints will be appreciated (if there is a site for counterexamples I will be glad).
 A: Here is an answer:
1) True. Since $\lim_{x\to \infty }xf'(x)=1,$ there exists $\alpha >0$ such that 
$$xf'(x)\geq \frac{1}{2}\; \forall\; x\geq \alpha.$$ In particular,
$$f'(x)\geq \frac{1}{2x}\; \forall\; x\geq \alpha.$$Now, if $x\geq \alpha,$ we have 
$$f(x)=f(\alpha)+\int_\alpha^x f'(t)dt\geq f(\alpha)+\int_\alpha^x \frac{1}{2x}dt$$ 
$$=f(\alpha)+ \frac{1}{2}\log(\frac{x}{\alpha})\to \infty.$$
2) False. Define $a_n= \frac{1}{n^2}$ if $n$ is not a perfect square and $a_n=\frac{1}{n}$ if $n$ is a perfect square. Then $\limsup na_n=1$ and $\sum a_n$ converges. On the other hand, if instead of  $\limsup na_n=1$ you impose  $\lim na_n=1,$ then the statement is true. You prove this as follows:
Since $\lim na_n=1,$ then exists $n_0$ such that for $n\geq n_0$ we have 
$$na_n\geq \frac{1}{2}, \textrm{ or equivalently, } a_n\geq \frac{1}{2n}.$$ The result follows from the fact that the series $\sum \frac{1}{n}$ diverges.
3) True. It suffices to prove that the sequence of partial sums,
$$S_n= \sum_{k=1}^n \max(a_k,a_{k+1})$$ is Cauchy. Since $\sum a_k$ converges, the sequence $$s_n= \sum_{k=1}^n a_k$$ is Cauchy. Now take any $\epsilon >0.$ Then, there exists $n_0$ such that $n\geq m\geq n_0$ implies 
$$s_n-s_m= \sum_{k=m}^n a_k \leq \frac{\epsilon}{2}.$$ Hence, for $n\geq m\geq n_0$ we also have 
$$S_n-S_m= \sum_{k=m}^n \max(a_k,a_{k+1})\leq 2 \sum_{k=m}^{n+1} a_k \leq 2\frac{\epsilon}{2}=\epsilon.$$ This means that $S_n$ is cauchy and hence convergent. 
Hope this helps
A: *

*Here I will just make a comment: Intuitively $\lim_{x\to\infty} xf'(x) = 1$ means that $f' \approx \frac1x$ for large $x$, so $f(x) \approx \ln x + C$ which grows to infinity as $x \to \infty$.

*Take $a_n = \frac{(-1)^n}{n}$. Then $n a_n = (-1)^n$ so $\limsup n a_n = +1$. The series $\sum_n a_n$ is alternating with $|a_n| \to 0$ and is therefore convergent.

*Set $b_n = a_n$ if $a_n \geq a_{n+1}$ and $=0$ otherwise. Set $c_n = a_{n+1}$ if $a_n < a_{n+1}$ and $=0$ otherwise. Then $0 \leq b_n \leq a_n$, $0 \leq c_n \leq a_{n+1}$ and $b_n+c_n = \max(a_n, a_{n+1})$. This implies that $0 \leq \sum b_n \leq \sum a_n < \infty$ and $0 \leq \sum c_n \leq \sum a_{n+1} < \infty$ so
$$\sum \max(a_n, a_{n+1}) = \sum (b_n+c_n) = \sum b_n + \sum c_n < \infty.$$
The splitting of the sum is okay when we only have non-negative terms.
