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Assume $a_n\geq 0$ is a sequence of positive real numbers which satisfy the following inequality: for each $n,m\in\mathbb{N}$, we have $$(n+m)a_{n+m}\leq na_n+ma_m.$$ I can't show the convergence of this seemingly well-behaved sequence (I am guessing it does converge, and that one should use a monotonic trick?). Any hint is appreciated. I found this in a list of exercises in sequences and convergence, and I believe it should be elementary because so are the rest of the problems in that list. Thanks in advance!

For completeness, my -failed- attempts so far:

  • Using induction, one proves esily that for any $p,n\in\mathbb{N}$ the inequality $$a_{pn}\leq a_n.$$ From here one might conclude convergence of subsequences of the form $\{a_{{p^k}n}\}_k$ for any $p,n\in\mathbb{N}$. But unfortunately, I can't get an argument from here.

  • Another promising-looking inequality is $$a_{n+1}\leq \frac{n}{n+1}a_n+\frac{a_1}{n},$$ but again this does not suffice.

  • Finally, I tried playing with the sequence (for some $r$ fixed) $$A_n=\min\left\{\frac{nx_n+x_1}{n+1},\frac{(n-1)x_{n-1}+x_2}{n+1},\dots,\frac{(n-1)x_{n-r}+x_{r+1}}{n+1}\right\},$$ becase its a trick I have seen elsewhere when solving exercises on convergence of bounded sequences. I don't think this is the way, because I believe this would only work if you had an estimate of $a_{n+1}$ as a "convex" combination of the $r$-tail.

EDIT: It seems like everyone can prove this result one way or another. A friend of mine immediately suggested to apply Fekete's Subadditive Lemma here to the sequence $b_n=n a_n$.

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This is a classic exercise, I give below a solution. I do not know if it is the simplest one. You have done a part of the work.

Put $L=\inf\{a_n\}$. Let $\varepsilon>0$.

a) There exists $m\geq 1$ such that $a_m<L+\varepsilon$. You have proven that then, for every $k\geq 1$, one has $a_{mk}\leq a_m\leq L+\varepsilon$.

b) Put $M=\max\{a_0,\cdots a_{m-1}\}$. Then for any $r\in \{0,\cdots,m-1\}$

$$a_{mk+r}\leq \frac{km}{km+r}a_{km}+\frac{r}{km+r}a_r\leq L+\varepsilon+\frac{M}{mk}$$

c) There exist $k_0$ such that $\frac{M}{mk}<\varepsilon$ for $k\geq k_0$.

d) Take $n\geq (k_0+1)m$. We can write $n=mk+r$, with $r\in \{0,\cdots,m-1\}$ and $k\geq k_0$. Hence by the above, $a_n\leq L+2\varepsilon$. As $a_n\geq L$, we are done.

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  • $\begingroup$ Thank you very much! Your solution looks really nice, I was even starting to hesitate whether $a_n$ was even converging. $\endgroup$ – Simulacra May 12 '17 at 13:47
  • $\begingroup$ You are welcome. $\endgroup$ – Kelenner May 12 '17 at 13:49

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