Convergence of $\sum \frac{1}{n}e_n$ in $\ell^p$ where $p\in[1,\infty]$ and $e_n$ is $(0,0,\dots,1,\dots)$

Question

Identify if $\sum \frac{1}{n}e_n$ converges in $\ell^p$ where $p\in[1,\infty]$ and $e_n$ is $(0,0,\dots,1,\dots)$ [it has 1 on $n$-th position, and 0 otherwise].

---

Now, I am kind of confused what should be done here. Are we working in product spaces? Unfortunately the exercise does not tell me that. Is it just a sequence with a "weird notation"?

Let me denote the partial sum $S_n = \sum_\limits{k=1}^nX_k = (1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}, 0, \dots)$

Now, do I have to check if sequence $x_n = S_n$ converges i.e. if $\lim x_n$ exists for some $p$ in $\ell^p$? I can also check if $\lim||x_n||$ exists because $\ell^p$ are Banach, and that makes absolute convergence iff we have regular convergence.

My attemp would be then:

• for $\ell^1$ the sequence diverges because $\lim_\limits{n \to \infty}||x_n||$ is just $\left(\sum_\limits{k=1}^\infty|\frac{1}{n}|^1\right)^1 = \infty$
• for $\ell^p$ and $p \in (1, \infty)$ the sequence converges because $\lim_\limits{n \to \infty}||x_n||$ = $\left(\sum_\limits{k=1}^\infty|\frac{1}{n}|^p\right)^{\frac{1}{p}} < \infty$
• for $\ell^{\infty}$ the supremum seems to be 1, so we have convergence

Is that correct reasoning? The product space idea makes me anxious.

Answer 1

Careful. Absolute convergence does not imply convergence of a sequence in a Banach space. Absolute convergence of a series implies the convergence of the series in a Banach space.

Let's distinguish those three cases, $p=1, p\in(1,\infty), p=\infty$. Every $\ell^p$ space is Banach so we have to check wether the sequence $(x_n)$ is Cauchy.

The sequence is $x_n=(1,1/2,\dots,1/n,0,0,0,\dots)$. Suppose that $n\leq m$.

If $p=1$, then $\|x_n-x_m\|_1=\sum_{k=n+1}^m\frac{1}{k}$ and this does not get small as the indexes n,m get large, because the series $\sum_{k=1}^\infty\frac{1}{l}$ diverges.

If $p\in(1,\infty)$ then $\|x_n-x_m\|_p^p=\sum_{k=n+1}^m\frac{1}{k^p}$ and this gets small as the indexes n,m get large, because the series $\sum_{k=1}^\infty\frac{1}{k^p}$ converges

If $p=\infty$ then $\|x_n-x_m\|_\infty=\frac{1}{n+1}$ and this gets small as the indexes n,m get large, so the series converges.