Suppose $a_{n}>0$ and the following series converges

$\sum_{n=1}^{\infty} a_{n}^{3}$

Does this imply that

$\sum_{n=1}^{\infty} \frac{a_{n}}{n}$


I was able to prove that the second series also converges by using the limit comparision test. Is there another way to show the second series converges (e.g. root or ratio test)?

  • $\begingroup$ If $a_n$ is a sequence of positive terms, one of Cauchy's tests might apply here: if $c_n$ is a term in your series, and $\lim_{n\rightarrow\infty}{\frac{\ln(1/c_n)}{\ln(n)}}<1$, your series is convergent. Applying this to the two series in the question, both get you something of the form $\lim_{n\rightarrow\infty}{\frac{\ln(a_n)}{\ln(n)}}$ with an additional additive or multiplicative term; thus if one converges, the other does as well. $\endgroup$ – J. M. is a poor mathematician Aug 6 '10 at 23:59
  • $\begingroup$ How did you get a limit comparison test to work here? For instance, take $a_n = \frac{1}{n}$. Then the first series is $\sum_{n=1}^{\infty} \frac{1}{n^3}$ and the second series is $\sum_{n=1}^{\infty} \frac{1}{n^2}$. Both converge, but the ratio of the successive terms of the second series to the first approaches infinity, so LCT does not apply. $\endgroup$ – Pete L. Clark Aug 7 '10 at 0:27
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    $\begingroup$ It is true for positive terms by an application of Minkowski's inequality: you can show that the partial sums of $\sum \frac{a_n}{n}$ are bounded. (This was mentioned in a deleted answer below.) $\endgroup$ – Akhil Mathew Aug 7 '10 at 1:02
  • $\begingroup$ @Pete. Consider 2 cases. Case 1: $a_{n}^{2} n$ converges to a non-zero number. Then the second series converges by the LCT. Case 2: $a_{n}^{2} n$ does not converge to a non-zero number. Break up $a_{n}$ into 2 subsequence $a_{k_{n}}$ and $a_{p_{n}}$ such that $a_{k_{n}}^{2} n \ge 1$ and $a_{p_{n}}^{2} n <1$ for all $n$. Then $\sum a_{k_{n}}/n$ converges by simple comparison test. Note that $a_{p_{n}}<1/ \sqrt{n}$ so $\sum a_{p_{n}}/n$ converges. So the second series converges under case 2 $\endgroup$ – Digital Gal Aug 7 '10 at 2:44
  • $\begingroup$ I would call that a little more than just LCT, but anyway it's a nice proof: thanks for giving it. Also your edit to clarify that the terms are non-negative is appreciated. $\endgroup$ – Pete L. Clark Aug 7 '10 at 3:15

$3|a_n|/n \leq (|a_n|^3 + (2/n^{3/2})$ by a well known inequality. Sum.

This also shows the range of values that the second infinite sum can assume, given the value of the first, and assuming all terms are non-negative.


Here's a more conceptual answer that doesn't resort to inequalities ad hoc. Assume $a_n > 0$.

The problem can be rephrased, by writing $(a_n)^3 = 1/(n^{3/2 + p})$ (with $p$ a function of $n > 1$, and ignoring $a_1$), as:

If $\Sigma 1/(n^{3/2 + p})$ converges then $\Sigma 1/(n^{3/2 + (p/3)})$ converges.

The transformation moves the series toward the (convergent) one with $p=0$. This preserves convergence, because the series can be split into the terms with $p \leq 0$ and $p>0$. For the first set, convergence is improved, and for the second set, the sum is dominated by $\Sigma 1/n^{3/2}$.

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    $\begingroup$ If one looks carefully at this proof, one sees that it relies on the not-so-conceptual argument that $a_n/n\leqslant a_n^3$ if $a_n\geqslant1/\sqrt{n}$ and $a_n/n\leqslant 1/n^{3/2}$ if $a_n\leqslant1/\sqrt{n}$. So much for the desire to avoid ad hoc inequalities... $\endgroup$ – Did Sep 19 '12 at 14:27

By Hölder's inequality you have $$ A_k=\sum_ {i=1}^{k} \frac{a_n}{n} \leq \left ( \sum_ {i=1}^{k}a_ n ^3 \right )^\frac{1}{3} \left ( \sum_ {i=1}^{k} \frac{1}{n^\frac{3}{2}}\right ) ^\frac{2}{3}$$ By taking $k \rightarrow \infty $,we see $\sum_ {i=1}^{\infty} \frac{a_n}{n}$ is bounded and since $\frac{a_n}{n} \geq 0$ we conclude it is convergent( since $A_k$ is bounded and increasing).


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