# How to prove that the sequence of $x_n = (1,\frac{1}{2}, \frac{1}{3}, … \frac{1}{n}, 0, 0…)$ does not converge under $\|\cdot\|_1$?

I'm reviewing past assignments and am still having trouble formulating a proof for this:

Consider the sequence $(x_n)$, where $x_n = (1,\frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n}, 0, 0, \ldots)$. Determine whether $(x_n)$ converges in $l_1$.

It's simple to show that the coordinate-wise limit does not converge, but how can I show that the coordinate-wise limit is the only possible limit? Alternatively, I'm trying to show that $(x_n)$ is unbounded, which I think should be straightforward also, but is giving me trouble.

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Are you aware of the Harmonic series? Then compare with the Basel problem – Henry Feb 24 '11 at 14:56
Quite aware; my difficulty is in formulating the proof; specificially that $x = (1,\frac{1}{2},...)$ is the only possible limit. – bosmacs Feb 24 '11 at 15:00

1. The standard trick for showing divergence is to estimate $\|x_{2n} - x_{n}\| = \|(0,\ldots,0,\frac{1}{n+1},\frac{1}{n+2},\ldots,\frac{1}{2n},0,\ldots)\| = \sum_{k=n+1}^{2n} \frac{1}{k} \geq n \cdot \frac{1}{2n} = \frac{1}{2},$ so the sequence $(x_{n})$ is not a Cauchy sequence, hence it can't converge. On the other hand, this also shows that $\|x_{2^k}\| \geq \frac{k}{2}$, so the sequence is unbounded.
2. Note that the map $\ell^{1} \to \mathbb{R}$ sending a sequence $(y_{n})_{n \in \mathbb{N}} \in \ell^{1}$ to its $n$th coordinate $y_{n}$ is continuous because $|y_{n}| \leq \|(y_{n})_{n \in \mathbb{N}}\|$. So an $\ell^{1}$-limit must be a pointwise limit. So if the pointwise limit is outside $\ell^{1}$, there can't be an $\ell^{1}$-limit.
Unboundedness should be straightforward: what is $||x_n||_1$? What does it do as $n \to \infty$?