Give an example which proves/disproves the condition for convergence of the series.

37 Views Asked by At

Firstly I have these 2 conditions for arbitrary sequences {$x_n$}:

Given any 1$\lt$q$\lt$p$\lt$ $\infty$, there is a real sequence {$x_n$} such that $\sum_{n=1}^{\infty}|{x_n}|^{p}$ is convergent but $\sum_{n=1}^{\infty}|{x_n}|^{q}$ is divergent.

I need an example to prove/disprove this. I tried to think of the auxiliary series $\sum {1\over n^p}$. But I ended up nowhere. A better example which I should consider here?

The second case is:

Given any 1$\lt$p$\lt$q$\lt$ $\infty$, there is a real sequence {$x_n$} such that $\sum_{n=1}^{\infty}|{x_n}|^{p}$ is convergent but $\sum_{n=1}^{\infty}|{x_n}|^{q}$ is divergent. I need an example to prove/disprove this as well.

1

There are 1 best solutions below

2
On BEST ANSWER

I see that in the comments the first case has already been answered. So, I'll focus on the second one.

The most important part here is to know how the tail of your series behaves. For example, look at $p=1$ and $q=2$. The tail should converge quicker of the series where $q=2$ if the terms are below $1$. For example, $1/n^2$ converges slower than $1/n^4$ to $0$.

To make this idea precise, we know that in a convergent series the terms should converge to $0$. Hence, we can find an $N\in \mathbb{N}$ such that for all $n\geq N$: $|x_n|<1$. And thus, the tail after $N$ is dominated by the $|x_n|^p$ series, since for all these $n$, we have $|x_n|^q < |x_n|^p$. Moreover, note that the convergence does not depend on the fixed finite first elements of the series $\{x_1,...,x_{N}\}$.

I hope this helps!