# Show $\sum_{k=1}^\infty|a_k|^q$ converges [duplicate]

Let $1 \leq p < q < \infty$. Show that if $x=(a_k)∈ℓ^p$

i.e. the condition that the series $$\sum_{k=1}^\infty|a_k|^p$$ converges holds, then $x∈ℓ^q$

i.e. $$\sum_{k=1}^\infty|a_k|^q$$ converges.

I did try to use the ratio test but I don't think it will work because of the power on the terms in the series.

## marked as duplicate by user296602, Did real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 27 '16 at 18:18

• What do $p$ and $q$ have to do with each other? – user296602 Jan 27 '16 at 17:46
Because $\sum_{n=1}^{\infty} |a_n|^p < \infty,$ $|a_n|^p \to 0.$ Thus there exists $N$ such that $|a_n|^p< 1$ for $n\ge N.$ For such $n$ we can say that, since $q/p > 1,$
$$|a_n|^q = (|a_n|^p)^{q/p}\le |a_n|^p.$$
Thus $\sum_{n=1}^{\infty} |a_n|^q$ converges by the comparison test.