# $0 \leq u_{n} \leq \frac {c}{n} \sum\limits_{k=1}^{n-1} u_{k}$ $\implies$ $\sum\limits_{k \geq 0} u_{k} < + \infty$?

I would be interesting in the question that whether we can relax the assumption $$0 \leq u_{n} \leq \frac {1}{n^{2}} \sum\limits_{k=1}^{n} u_{k}$$ to $$0 \leq u_{n} \leq \frac {c}{n} \sum\limits_{k=1}^{n-1} u_{k}$$ for some constant $$c \in \left( 0 , 1 \right)$$. Note that we also has one term less than the previous (the series only run till $$u_{n-1}$$).

We can not use the same approach as in the previous question since $$\sum\limits_{k \geq 1} u_{k} = + \infty$$.

• Are you willing to ignore $n=1$ in the hypothesis? May 8 '19 at 7:23

With the convention that an empty sum evaluates to zero, we have for $$n=1$$ $$0 \le u_1 \le \frac c1 \sum_{k=1}^0 u_k = 0$$ which implies $$u_1= 0$$. Then $$u_n= 0$$ for all $$n\in \Bbb N$$ follows via induction.

So the only sequence satisfying the hypothesis is the zero sequence, so that $$\sum_{k \geq 0} u_{k}$$ is (trivially) finite.

If the hypothesis is relaxed to $$\tag{*} 0 \leq u_{n} \leq \frac {c}{n} \sum_{k=1}^{n-1} u_{k} \text{ for sufficiently large n}$$ then for any $$c \in (0, 1)$$ a counter-example is given by $$u_n = (n+1)^d - n^d$$ where $$d$$ is chosen in $$(0, c)$$. From the mean-value theorem we can estimate the left-hand side as $$u_n = (n+1)^d - n^d \le d n^{d-1}$$ so that $$(*)$$ is satisfied if $$d n^{d-1} \le \frac {c}{n} \sum_{k=1}^{n-1} u_{k} = \frac {c}{n} \bigl( n^d - 1\bigr) \\ \iff c \le (c-d)n^d$$ and that is true for sufficiently large $$n$$.

• +1 from me. Sorry about earlier correspondence. May 8 '19 at 11:44

I am ignoring first few values of $$n$$ in the hypothesis. $$u_n=\log(1+\frac 1 {n+2})$$ is a counterexample. Note that $$\sum\limits_{k=1}^{n-1} u_k=\sum\limits_{k=1}^{n-1}[log(k+3)-log(k+2)]=\log(n+2)-\log\, 3$$ from which you can see that the hypothesis is satisfied.

• Note that for $n=2$, we get in the given example $u_1=\ln(\frac43) \approx 0.288$ and $u_2=\ln(\frac54) \approx 0.223$ and thus $\frac1n\sum_{k=1}^{n-1}=\frac12u_1 \approx 0.144$. That means the constant $c$ cannot be smaller than $\approx 1.5$, which is obviusly outside the specfied range. May 8 '19 at 10:03
• Since the conditions are linear in $\{u_n\}$, one cannot get a lower $c$ by just multiplying $\{u_n\}$ with a constant. Even if the denominator $n$ is replaced by the more 'appropriate' $n-1$ (it is, after all, dividing a sum of $n-1$ values), that only covers values of $c \ge \approx 0.75$. It might still be interesting to know if the assumption is true for some 'small enough' $c > 0$. May 8 '19 at 10:03

Ignoring the condition for $$n=1$$, the assumption is incorrect for any $$c > 0$$.

To see this, define $$u_1:=1$$ and

$$u_n:=\frac{c}n\sum_{k=1}^{n-1}u_k, \text{ for } n\ge 2.$$

(REMARK: See below for a simplification of this proof).

This satisfies the condition of the problem, we have greedily chosen $$u_n$$ to be the maximal available value.

It follows for $$n \ge 2$$ that $$u_n \ge \frac{c}n u_1 = \frac{c}n$$, as the remaining terms of the sum are non-negative.

Thus we get

$$\sum_{k=1}^nu_k \ge \sum_{k=2}^nu_k \ge \sum_{k=2}^n\frac{c}k = c\sum_{k=2}^n\frac1k.$$ The sum on the right hand side is the known divergent harmonic series, and with $$c > 0$$ our sum is divergent by the comparison test.

SIMPLIFICATION:

After $$u_1:=1$$, simpy continue with the explicit definition

$$u_n:=\frac{c}n, \text{ for } n\ge 2.$$

Since $$u_n=\frac{c}nu_1$$, the conditions is fullfilled for all $$n\ge 2$$ and divergence follows the same way as above.

• thanks for your observation. so what do you think about the case $0 \leq u_{n} \leq \dfrac{c}{n^{p}} \sum_{k=1}^{n−1} u_{k}$ for $1<p<2$? I took the power $1$ since I thought it would be easier in the analysis.
– JKay
May 9 '19 at 1:29