If $\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} f_{n}(k) = \infty$ then $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} f_{n}(k) = \infty$.

Suppose $ \{f_{n}\}_{n=1}^{\infty}$ be functions such that $ f_{n} : \Bbb{N} \rightarrow \Bbb{R}^{+}$ for each $ n$ .

I was trying to prove –

If $ \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} f_{n}(k) = \infty$ then $ \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} f_{n}(k) = \infty$ .

I can see that both the double summation series are equal by expanding the double summation.But I am trying to prove it ? any other thoughts?

Hm, as the order of summation are changed, is Uniform convergence likely to play any role here?

Find all the $c \ge 0$ for which $\sum_{n=1}^{+ \infty }a _{n}$ is absolutely convergent

We consider: $ $ a_{1}=c-1$ $ $ $ a_{n+1}= \frac{-n}{n+c \cdot \sqrt[n]{ln(n^{9876}+17)}}\cdot a_{n}, n\ge 1$ $ $ $ c\ge0$ $ I want to use Rabbe Test, because then in a simple way it comes out that the series is convergent for every $ c\ge0$ . However I have two doubts: 1) Raabe Test is for $ a_{n}>0$ , but if I do $ r_{n}=n(|\frac{a_{n}}{a_{n+1}}|-1)$ I knew that I must removing minus at $ n$ and then I can leave the module. Hovewer I’m not sure if it’s allowed. 2) If Raabe Test is a good way to do this task I knew only when my series is convergent, but I don’t knew when is absolutely convergent so the more I do not know if Raabe Test is a good idea.

From $lim_{x \to +\infty} \frac{f(x)}{x^p}$ to $lim_{x \to +\infty} \frac{f'(x)}{px^{p-1}}$?

Define $ f: (0 , +\infty)\to \mathbb{R}$ differentiable satisfying $ \displaystyle\lim_{x \to +\infty} \frac{f(x)}{x^p} = 1$ and $ f’$ is monotonically increasing.

Can we conclude $ \displaystyle\lim_{x \to +\infty} \frac{f'(x)}{px^{p-1}}=1$ ?

My try:

Attempt 1:(try to use Lagrange Mean Value Theorem)

Considering $ \displaystyle\lim_{x \to +\infty} \frac{f(x+1)-f(x)}{x^{p-1}} $ ,where $ f(x+1)-f(x)=f'(\zeta)$ , $ x<\zeta<x+1$ \begin{align} \frac{f(x+1)-f(x)}{x^{p-1}} &=\frac{f(x+1)}{(x+1)^p}\frac{(x+1)^p}{x^p}x-\frac{f(x)}{x^p}x \&=\frac{f(x+1)}{(x+1)^p}\frac{x^{p+1}+px^p+\mathbb{o}(x^p)}{x^p}-\frac{f(x)}{x^p}x \&=\frac{f(x+1)}{(x+1)^p}(x+p+\mathbb{o}\left(1\right))-\frac{f(x)}{x^p}x \&=p\frac{f(x+1)}{(x+1)^p}+x(\frac{f(x+1)}{(x+1)^p}-\frac{f(x)}{x^p})+\mathbb{o}\left(1\right) \end{align} Because $ f’$ is monotonically increasing we just need to prove $ \displaystyle\lim_{x \to +\infty} \frac{f(x+1)-f(x)}{x^{p-1}}=p $ then we can arrive at $ \displaystyle\lim_{x \to +\infty} \frac{f'(x)}{px^{p-1}}=1$

So I aimed to prove $ \displaystyle\lim_{x \to +\infty} x(\frac{f(x+1)}{(x+1)^p}-\frac{f(x)}{x^p})=0$ .

But I failed and didn’t know how to move on…

Attempt 2:(try to use Cauchy Mean Value Theorem)

Considering $ \displaystyle\lim_{x \to +\infty} \frac{f(x+1)-f(x)}{(x+1)^p-x^p} $ , where $ \dfrac{f(x+1)-f(x)}{(x+1)^p-x^p}=\dfrac{f'(\xi)}{p \xi^{p-1}}$ , $ x<\xi<x+1$

Similarly , because $ f’$ is monotonically increasing we just need to prove $ \displaystyle\lim_{x \to +\infty} \frac{f(x+1)-f(x)}{(x+1)^p-x^p} =1 $ then we can arrive at $ \displaystyle\lim_{x \to +\infty} \frac{f'(x)}{px^{p-1}}=1$

Actually I found it equivalent to proving that $ \displaystyle\lim_{x \to +\infty} \frac{f(x+1)-f(x)}{x^{p-1}}=p $ $ \ldots$

Any hints ? Thank you in advance!