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Assume $a_n > 0$. Define $r = \lim(\frac{\sqrt{a_n a_{n+1}}}{a_n}) = \lim(\sqrt{\frac{a_{n+1}}{a_n}}) = 1$ since $a_n$ is convergent. So by limit comparison test $\sum{\sqrt{a_n a_{n+1}}}$ is convergent.

Most proofs I've seen on this use AM-GM inequality so I thought I'd ask about this one.

Darby Bond
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    $\lim_{n \to\infty} \sqrt{\frac{a_{n+1}}{a_{n}}} = 1$ need not be true, if $a_{n} = e^{-n}$ for example. This case is not a problem though. However, the ratio test allows for the possibility of the limit being zero where the comparison test (in its usual form) does not apply. I don't know immediately if it is easy to fix. – Manifoldski Oct 16 '22 at 00:53
  • I suppose a reasonable question is if we know that $a_n$ converges to $0$ (as it will if $\sum{a_n}$ converges), does $\frac{a_{n+1}}{a_n}$ converge? – Darby Bond Oct 16 '22 at 01:09
  • Unless I am mistaken (its late here) this tells you that $\frac{a_{n+1}}{a_{n}}$ is not necessarily convergent under the assumption of convergence of the series. – Manifoldski Oct 16 '22 at 01:15

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In your proof attempt, you make the claim that if $$ \sum_{n=1}^\infty a_n $$ is a convergent series of positive terms, then $$ \lim_{n\to\infty}\sqrt{\frac{a_{n+1}}{a_n}}=1 $$ but that's an invalid claim.

In fact, that limit may not even exist.

For example, consider the series $\sum a_n$ where $a_n={\large{\frac{1}{2^n}}}$ when $n$ is odd, and $a_n=2a_{n-1}$ when $n$ is even.

quasi
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