The Cauchy test states $\sum a_n$ converges $\iff$ $\sum 2^k a_{2^k}$ converges.

In Rudin 3ed, the (excerpt) proof is outlined as follows

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Can I modify (8) so we get

For $n < 3^k$

$$s_n = (a_1 + a_2 + a_3) + (a_4 + \dots + a_9) \leq 3a_1 + 3^2a_3 + \dots + 3^{k+1} a_{3^k} =t_k$$

For $n > 3^k$

$$s_n = (a_1 + a_2) + (a_3 + a_4 + \dots + a_{11} ) + a_{12} + \dots \geq a_1 + 2a_1 + 3^2 a_3 + \dots = t_k$$

So that we may test $\sum 3^{k+1}a_{3^k}$ instead?


The calculation can be changed to use $b^k a_{b^k}$ for any integer $b\ge 1$. The proof, like the usual proof for the case $b=2$, is by grouping, and finding upper and lower bounds an upper bound for the sum of the $a_i$ from $i=b^{k}+1$ to $i=b^{k+1}$ (for the convergence part).

Of course $b^{k+1}a_{b^k}$ makes no difference. And we can iterate and use very much faster-growing indices.

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  • $\begingroup$ What about $f(n) a_{f(n)}$ - what conditions must $f$ satisfy to have $\sum a_n < \infty \iff \sum f(n) a_{f(n)} < \infty$ ? Where could I find some more info on this? $\endgroup$ – Spine Feast Aug 31 '13 at 15:12
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    $\begingroup$ I don't have references. There is a more general version called the Schlomilch (sp?) test that might get you started. $\endgroup$ – André Nicolas Aug 31 '13 at 15:23

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