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If $1 \leq q < p$ whats a sequence in $\ell^{p}$ but not in $\ell^{q}$?

Having trouble understanding this, just looking for an example. Thank you.

1 Answers 1

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The sequence $x_n=n^{-\frac{1}{q}}$ will do.

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    $x = \left\{x_n \right\}$ is not in $\ell^q$ because $\|\ {x} \|_{\ell^{q}}\ = \left( \sum_{n=1}^{\infty} |n^{- \frac{1}{q}}|^{q} \right)^{\frac{1}{q}} = \left( \sum_{n=1}^{\infty} |\frac{1}{n}| \right)^{\frac{1}{q}}$ because this diverges? – 2017-02-04
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    Yes, the harmonic series diverges. – 2017-02-04
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    and from above, this is in $\ell^{p}$ because $\|\ {x} \|_{\ell^{p}}\ = \left( \sum_{n=1}^{\infty} |n^{- \frac{1}{q}}|^{p} \right)^{\frac{1}{p}} = \left( \sum_{n=1}^{\infty} |\frac{1}{n}|^{\frac{p}{q}} \right)^{\frac{1}{p}}$ because this converges? – 2017-02-04
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    because $\frac{p}{q} > 1$ so by $p$-test converges. – 2017-02-04
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    the fact that the whole series is raised to the $\frac{1}{q}$ doesn't effect the divergence for $\|\ x \|_{\ell^{q}}\ $? – 2017-02-04
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    This doesn't matter. – 2017-02-04
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    Okay, thank you. If possible would you elaborate why that doesn't matter? It's not clear to me (albeit probably should be). – 2017-02-04
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    Because infinity raise to any positive power is still infinity. – 2017-02-04