1-Sequence Space is Proper Subset of 2-Sequence Space

Theorem
Let $\ell^1$ and $\ell^2$ be the 1-sequence space and 2-sequence space respectively.

Then $\ell^1$ is a proper subset of $\ell^2$.

$\ell^1$ and $\ell^2$ share some elements
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\ell^1$.

By definition:


 * $\ds \sum_{n \mathop = 0}^\infty \size {x_n} < \infty$

By Terms in Convergent Series Converge to Zero:


 * $\ds \lim_{n \mathop \to \infty} \size {x_n} = 0$

By definition of convergent sequence:


 * $\exists N \in \N : \forall n > N : \size {x_n} < 1$

Furthermore:


 * $\forall \size {x_n} < 1 : \size {x_n}^2 < \size {x_n}$

By comparison test:


 * $\ds \sum_{n \mathop = 0}^\infty \size {x_n}^2 < \infty$

Hence, $\sequence {x_n}_{n \mathop \in \N} \in \ell^2$

$\ell^1$ and $\ell^2$ are not equinumerous
Let $\ds x := \tuple {1, \frac 1 2, \ldots, \frac 1 n, \ldots}$ with $n \in \N_{\mathop > 0}$.

From Basel Problem:


 * $\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} < \infty$

From Harmonic Series is Divergent:


 * $\ds \sum_{n \mathop = 1}^\infty \frac 1 n$

diverges.

Hence, $x \in \ell^2$, but $x \notin \ell^1$.

Therefore, $\ell^1 \subsetneqq \ell^2$.