Let $\{ a_n \}_{n=1}^{\infty}$ be a sequence of non-negative real numbers. If $\sum_{n=1}^{\infty}a_n = \infty$, is it also true that $$ \sum_{n=1}^{\infty}\frac{a_n}{1+\sum_{k=1}^{n}a_k}=\infty? $$ Note: This is not a homework problem or something I found in a text. Just a curious way of constructing a more slowly divergent series from a divergent series. Maybe it's a standard construction. I don't know.

What I tried: This is related to log, and a valid argument in that vein seems to work for a bounded sequence $\{ a_n \}$ by considering the telescoping sum $$ \sum_{n=1}^{\infty}\ln\left(1+\frac{a_{n+1}}{1+\sum_{k=1}^{n}a_k}\right)=\sum_{n=1}^{\infty}\left[\ln\left(1+\sum_{k=1}^{n+1}a_k\right)-\ln\left(1+\sum_{k=1}^{n}a_k\right)\right] = \infty $$ I don't think I can deal with an unbounded sequence using this argument because of how the index in the numerator on the far left is off by 1 from the denominator.

  • $\begingroup$ Is it $a_n$ or $a_{n+1}$ in the numerator (in the two formulas you have two different choices)? $\endgroup$ – m7e Apr 25 '16 at 20:31
  • $\begingroup$ @mge : That's why the formula I want I can only get for bounded sequences. What I want is the formula as stated in the problem statement. $\endgroup$ – DisintegratingByParts Apr 25 '16 at 20:39

Let $S_n = \sum_{k=1}^n a_k$. This sequence is increasing, as $a_n$ is non-negative, and diverges to $+\infty$ by hypothesis.

Then for any fixed $n$,

$$\lim_{m \to \infty} \frac{1 + S_n}{1 + S_m} = 0.$$

If $m > n$ is sufficiently large then

$$\frac{1 + S_n}{1 + S_m} < 1/2,$$

and $$\left|\sum_{k = n+1}^m \frac{a_k}{1+S_k}\right| \geqslant\frac{1+S_m - (1+S_n)}{1+S_m} = 1 - (1+S_n)/(1+S_m) > 1/2$$

Therefore, the series $\sum a_n/(1+S_n)$ diverges to $+\infty$ since the Cauchy criterion is violated and the terms are non-negative.

  • $\begingroup$ Very nice. Thank you. I never would have found your solution. :) $\endgroup$ – DisintegratingByParts Apr 25 '16 at 21:25
  • 1
    $\begingroup$ @TrialAndError: You're welcome. Generally harder to prove divergence directly with no obvious comparison. $\endgroup$ – RRL Apr 25 '16 at 21:38

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