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Let $\sum u_n$ and $\sum v_n$ two series. Assume for all n $u_n,v_n>0$ and $\sum v_n$ is finite and $$ \frac{u_{n+2}}{u_n} \le \frac{v_{n+2}}{v_n}\qquad \text{ for all n }\in \mathbb{N} $$ Show that $\sum u_n$ is finite.


Define $w_n = u_{2n}$ and $a_n=v_{2n}$ Since $v_n>0$ we know that $\sum a_n$ is finite.

Applying the logarithm and summing, we have $$\sum_{n=0}^{p-1}(\ln(w_{n+1})-\ln(w_n))\le \sum_{n=0}^{p-1}(\ln(a_{n+1})-\ln(a_n)) = \ln(a_p)-\ln(a_0)$$

Hence $$w_p \le k\,a_p \qquad \text{ for all p }\in \mathbb{N}$$ where $k=\exp(\ln(w_0)-\ln(a_0))$. Hence $\sum w_n$ is finite.

I can do the same with $u_{2n+1}$ and so by sum of finite series $\sum u_n$ is finite.

Do you have any remark and/or other solutions ?

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    $\begingroup$ Seems fine to me. $\endgroup$ – Kavi Rama Murthy Aug 14 at 8:00
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An other answer: write $$\frac{w_{n+1}}{w_n}\le \frac{a_{n+1}}{a_n}$$

Fix M>0. The sequence $(\frac{w_n}{a_n})_n$ is descreasing so there is a integer N such that $\frac{w_n}{a_n}< M$ for all $n>N$.

Then for all $n>N$, $w_{n+1}\le \frac{w_{n}}{a_{n}}a_{n+1}\le M a_{n+1}$

So $\sum w_n$ is finite. Like $\sum u_{2n+1}$ is finite, hence $\sum u_n$ is finite.

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  • $\begingroup$ this is wrong since $N$ may not exist .... $\endgroup$ – Smilia Sep 2 at 13:07

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