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Iosif Pinelis
Iosif Pinelis
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The answer is no. E.g., let $a_m:=1/(m!)^2$ and $n_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na_{m_n}|$ is bounded. In our case, we have $$a_{k+1}<\frac1{n_k}<a_k$$ for all $k$. So, for all $m\le k$ we have $1<n_k a_k\le n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$$|\ln(n_k a_m)|\ge|\ln(n_k a_k)|=\ln(k+1)$. Similarly, for all $m\ge k+1$ we have $1>n_k a_{k+1}\ge n_k a_m$ and hence again $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$$|\ln(n_k a_m)|\ge|\ln(n_k a_{k+1})|=\ln(k+1)$. So, $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$ for$|\ln(n_k a_m)|\ge\ln(k+1)$ for all $k,m$.

So, for any $(m_n)$ $$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\min(|\ln n_k a_k|,|\ln n_k a_{k+1}|)=\ln(k+1)\to\infty$$$$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\ln(k+1)\to\infty$$ as $k\to\infty$, so that $|\ln na_{m_n}|$ is unbounded, for any choice of $(m_n)$.

The answer is no. E.g., let $a_m:=1/(m!)^2$ and $n_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na_{m_n}|$ is bounded. In our case, we have $$a_{k+1}<\frac1{n_k}<a_k$$ for all $k$. So, for all $m\le k$ we have $1<n_k a_k\le n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$. Similarly, for all $m\ge k+1$ we have $1>n_k a_{k+1}\ge n_k a_m$ and hence again $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$. So, $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$ for all $k,m$.

So, for any $(m_n)$ $$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\min(|\ln n_k a_k|,|\ln n_k a_{k+1}|)=\ln(k+1)\to\infty$$ as $k\to\infty$, so that $|\ln na_{m_n}|$ is unbounded, for any choice of $(m_n)$.

The answer is no. E.g., let $a_m:=1/(m!)^2$ and $n_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na_{m_n}|$ is bounded. In our case, we have $$a_{k+1}<\frac1{n_k}<a_k$$ for all $k$. So, for all $m\le k$ we have $1<n_k a_k\le n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|=\ln(k+1)$. Similarly, for all $m\ge k+1$ we have $1>n_k a_{k+1}\ge n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_{k+1})|=\ln(k+1)$. So, $|\ln(n_k a_m)|\ge\ln(k+1)$ for all $k,m$.

So, for any $(m_n)$ $$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\ln(k+1)\to\infty$$ as $k\to\infty$, so that $|\ln na_{m_n}|$ is unbounded, for any choice of $(m_n)$.

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

The answer is no. E.g., let $a_m:=1/(m!)^2$ and $n_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na_{m_n}|$ is bounded. In our case, we have $$a_{k+1}<\frac1{n_k}<a_k$$ for all $k$. So, for all $m\le k$ we have $1<n_k a_k\le n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$. Similarly, for all $m\ge k+1$ we have $1>n_k a_{k+1}\ge n_k a_m$ and hence again $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$. So, $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$ for all $k,m$.

So, for any $(m_n)$ $$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\min(|\ln n_k a_k|,|\ln n_k a_{k+1}|)=\ln(k+1)\to\infty$$ as $k\to\infty$, so that $|\ln na_{m_n}|$ is unbounded, for any choice of $(m_n)$.

The answer is no. E.g., let $a_m:=1/(m!)^2$ and $n_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na_{m_n}|$ is bounded. In our case, we have $$a_{k+1}<\frac1{n_k}<a_k$$ for all $k$ and hence for any $(m_n)$ $$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\min(|\ln n_k a_k|,|\ln n_k a_{k+1}|)=\ln(k+1)\to\infty$$ as $k\to\infty$, so that $|\ln na_{m_n}|$ is unbounded, for any choice of $(m_n)$.

The answer is no. E.g., let $a_m:=1/(m!)^2$ and $n_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na_{m_n}|$ is bounded. In our case, we have $$a_{k+1}<\frac1{n_k}<a_k$$ for all $k$. So, for all $m\le k$ we have $1<n_k a_k\le n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$. Similarly, for all $m\ge k+1$ we have $1>n_k a_{k+1}\ge n_k a_m$ and hence again $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$. So, $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|$ for all $k,m$.

So, for any $(m_n)$ $$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\min(|\ln n_k a_k|,|\ln n_k a_{k+1}|)=\ln(k+1)\to\infty$$ as $k\to\infty$, so that $|\ln na_{m_n}|$ is unbounded, for any choice of $(m_n)$.

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

The answer is no. E.g., let $a_m:=1/(m!)^2$ and $n_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na_{m_n}|$ is bounded. In our case, we have $$a_{k+1}<\frac1{n_k}<a_k$$ for all $k$ and hence for any $(m_n)$ $$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\min(|\ln n_k a_k|,|\ln n_k a_{k+1}|)=\ln(k+1)\to\infty$$ as $k\to\infty$, so that $|\ln na_{m_n}|$ is unbounded, for any choice of $(m_n)$.