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I have read that :

If $\sum_{n=1}^{\infty}a_n$ is a series whose terms form a positive monotone decreasing sequence $(a_n)$, then it converges and diverges with $$ \sum_{k=0}^{\infty}2^ka_{2^k} = a_1 + 2a_2 + 4a_4 + 8a_8 + \cdots $$ My question now is : If conditions for $(a_n)$ hold as above, can we argue the same for the subsequence below $$ \sum_{k=0}^{\infty}3^ka_{3^k} = a_1 + 3a_3 + 9a_9 + 27a_{27} + \cdots $$ or more general for a subsequence $$ \sum_{k=0}^{\infty}n^ka_{n^k} = a_1 + na_n + n^2a_{n^2} + n^3a_{n^3} + \cdots $$

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Yes, the original case using powers of $n = 2$ is typically called the Cauchy condensation test. However, you can use any integer $n \geq 2$ and show that your original series is bounded below by your given shortened series and bounded above by $n$ times that lower bound. So the "if and only if" of Cauchy condensation test works with any choice of multiplier $n \geq 2$, not just $n=2$.

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For short: yes, the Cauchy's condensation test still works if we replace the increasing function $g(k)=2^k$ with any increasing function that fulfills:

$$ A\cdot g(k)\leq g(k+1)-g(k) \leq B\cdot g(k+1) \tag{1} $$ with $A,B$ being two positive constants. It is enough to check that the proof on Wikipedia goes along exactly the same lines if we assume just $(1)$ instead of $g(k)=2^k$.

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  • $\begingroup$ Thanks for the insight Jack. Nice generalization. $\endgroup$ – nickchalkida Sep 11 '15 at 20:51

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