# Show that $\sum_{n=1}^{\infty}a_n$ converges if $\sum_{k=0}^{\infty}2^ka_{2^k}$ converges… [duplicate]

Let $(a_n)_{n\in\mathbb{N}}\in[0,\infty[^{\mathbb{N}}$ be a monotonically decreasing sequence, so that $\sum_{k=0}^{\infty}2^ka_{2^k}$ converges. Show that $\sum_{n=1}^{\infty}a_n$ converges and test $\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}}$ for convergence.

Can anybody…explain to me what this exercise even asks of me? I have trouble understanding it in the first place. Is $a_n=2^ka_{2^k}$ here or something else? If so am I simply supposed to show that $\sum_{n=1}^{\infty}2^{n}a_{2^{n}}$ converges? Seems unlikely. I don't get it.

## marked as duplicate by Robert Z calculus 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 5 '18 at 22:21

Observe $$\sum_{n=1}^\infty a_n = \sum_{k=0}^\infty \sum_{n=2^{k}}^{2^{k+1}-1} a_n \le \sum_{k=0}^\infty 2^{k} a_{2^{k}} < \infty.$$ So you group the elements of the original sequence into exponentially larger blocks, and use monotonicity in each block.