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In this question it is proved that: $0\leq u_n\leq \frac {1}{n^2}\sum_{k=1}^nu_k\implies $ $\sum u_k$ converges.

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 $\sum\limits_{k \geq 1} u_{k} = + \infty$.

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

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

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    $\begingroup$ 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. $\endgroup$
    – Ingix
    May 8 '19 at 10:03
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    $\begingroup$ 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' $c > 0$. $\endgroup$
    – Ingix
    May 8 '19 at 10:03
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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.

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  • $\begingroup$ 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. $\endgroup$
    – JKay
    May 9 '19 at 1:29

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