# Series and Finitness — Quotient of close terms bounded by close terms of finite series

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 ?

• Seems fine to me. – Kavi Rama Murthy Aug 14 at 8:00

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