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edit approved Sep 20, 2021 at 14:08
Goulifet
edit approved Sep 20, 2021 at 14:08
Goulifet
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$\newcommand\ep\epsilon$The answer is no.

Indeed, let e.g. $u_k:=1$ if $k=k_j:=2^{5^j}$ for some natural $j$, and $u_k:=0$ otherwise. Then $p_c=0$.

Take now any real $\ep>0$ and then take $p=2\ep$, so that condition $0<p<p-p_c$$0< \epsilon <p-p_c$ holds. Then for all large enough $j$ and $n=k_{j+1}-1$ we have $$n^{p-p_c-\ep}=(k_{j+1}-1)^\ep\ge2^{\ep 5^{j+1}/2}$$ and $$\sum_{k=1}^n k^pu_k=\sum_{i=1}^j k_i^{2\ep} \le2k_j^{2\ep}=2\times2^{2\ep 5^j} =o(2^{\ep 5^{j+1}/2})=o(n^{p-p_c-\ep})$$ as $j\to\infty$.

So, there is no real $m>0$ such that $mn^{p-p_c-\ep}\le\sum_{k=1}^n k^pu_k$ for all $n$.

$\newcommand\ep\epsilon$The answer is no.

Indeed, let e.g. $u_k:=1$ if $k=k_j:=2^{5^j}$ for some natural $j$, and $u_k:=0$ otherwise. Then $p_c=0$.

Take now any real $\ep>0$ and then take $p=2\ep$, so that condition $0<p<p-p_c$ holds. Then for all large enough $j$ and $n=k_{j+1}-1$ we have $$n^{p-p_c-\ep}=(k_{j+1}-1)^\ep\ge2^{\ep 5^{j+1}/2}$$ and $$\sum_{k=1}^n k^pu_k=\sum_{i=1}^j k_i^{2\ep} \le2k_j^{2\ep}=2\times2^{2\ep 5^j} =o(2^{\ep 5^{j+1}/2})=o(n^{p-p_c-\ep})$$ as $j\to\infty$.

So, there is no real $m>0$ such that $mn^{p-p_c-\ep}\le\sum_{k=1}^n k^pu_k$ for all $n$.

$\newcommand\ep\epsilon$The answer is no.

Indeed, let e.g. $u_k:=1$ if $k=k_j:=2^{5^j}$ for some natural $j$, and $u_k:=0$ otherwise. Then $p_c=0$.

Take now any real $\ep>0$ and then take $p=2\ep$, so that condition $0< \epsilon <p-p_c$ holds. Then for all large enough $j$ and $n=k_{j+1}-1$ we have $$n^{p-p_c-\ep}=(k_{j+1}-1)^\ep\ge2^{\ep 5^{j+1}/2}$$ and $$\sum_{k=1}^n k^pu_k=\sum_{i=1}^j k_i^{2\ep} \le2k_j^{2\ep}=2\times2^{2\ep 5^j} =o(2^{\ep 5^{j+1}/2})=o(n^{p-p_c-\ep})$$ as $j\to\infty$.

So, there is no real $m>0$ such that $mn^{p-p_c-\ep}\le\sum_{k=1}^n k^pu_k$ for all $n$.

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Iosif Pinelis
Iosif Pinelis
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$\newcommand\ep\epsilon$The answer is no.

Indeed, let e.g. $u_k:=1$ if $k=k_j:=2^{5^j}$ for some natural $j$, and $u_k:=0$ otherwise. Then $p_c=0$.

Take now any real $\ep>0$ and then take $p=2\ep$, so that condition $0<p<p-p_c$ holds. Then for all large enough $j$ and $n=k_{j+1}-1$ we have $$n^{p-p_c-\ep}=(k_{j+1}-1)^\ep\ge2^{\ep 5^{j+1}/2}$$ and $$\sum_{k=1}^n k^pu_k=\sum_{i=1}^j k_i^{2\ep} \le2k_j^{2\ep}=2\times2^{2\ep 5^j} =o(2^{\ep 5^{j+1}/2})=o(n^{p-p_c-\ep})$$ as $j\to\infty$.

So, there is no real $m>0$ such that $mn^{p-p_c-\ep}\le\sum_{k=1}^n k^pu_k$ for all $n$.

$\newcommand\ep\epsilon$The answer is no.

Indeed, let e.g. $u_k:=1$ if $k=k_j:=2^{5^j}$ for some natural $j$, and $u_k:=0$ otherwise. Then $p_c=0$.

Take now any real $\ep>0$ and then take $p=2\ep$, so that condition $0<p<p-p_c$ holds. Then for all large enough $j$ and $n=k_{j+1}-1$ we have $$n^{p-p_c-\ep}=(k_{j+1}-1)^\ep\ge2^{\ep 5^{j+1}/2}$$ and $$\sum_{k=1}^n k^pu_k=\sum_{i=1}^j k_i^{2\ep} \le2k_j^{2\ep}=2\times2^{2\ep 5^j} =o(2^{\ep 5^{j+1}/2})=o(n^{p-p_c-\ep})$$ as $j\to\infty$.

$\newcommand\ep\epsilon$The answer is no.

Indeed, let e.g. $u_k:=1$ if $k=k_j:=2^{5^j}$ for some natural $j$, and $u_k:=0$ otherwise. Then $p_c=0$.

Take now any real $\ep>0$ and then take $p=2\ep$, so that condition $0<p<p-p_c$ holds. Then for all large enough $j$ and $n=k_{j+1}-1$ we have $$n^{p-p_c-\ep}=(k_{j+1}-1)^\ep\ge2^{\ep 5^{j+1}/2}$$ and $$\sum_{k=1}^n k^pu_k=\sum_{i=1}^j k_i^{2\ep} \le2k_j^{2\ep}=2\times2^{2\ep 5^j} =o(2^{\ep 5^{j+1}/2})=o(n^{p-p_c-\ep})$$ as $j\to\infty$.

So, there is no real $m>0$ such that $mn^{p-p_c-\ep}\le\sum_{k=1}^n k^pu_k$ for all $n$.

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Iosif Pinelis
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

$\newcommand\ep\epsilon$The answer is no.

Indeed, let e.g. $u_k:=1$ if $k=k_j:=2^{5^j}$ for some natural $j$, and $u_k:=0$ otherwise. Then $p_c=0$.

Take now any real $\ep>0$ and then take $p=2\ep$, so that condition $0<p<p-p_c$ holds. Then for all large enough $j$ and $n=k_{j+1}-1$ we have $$n^{p-p_c-\ep}=(k_{j+1}-1)^\ep\ge2^{\ep 5^{j+1}/2}$$ and $$\sum_{k=1}^n k^pu_k=\sum_{i=1}^j k_i^{2\ep} \le2k_j^{2\ep}=2\times2^{2\ep 5^j} =o(2^{\ep 5^{j+1}/2})=o(n^{p-p_c-\ep})$$ as $j\to\infty$.