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Terry Tao
Terry Tao
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No, one can create sequences in which the first sequence converges but the second does not (or vice versa).

For sake of argument take $k=2$. To begin with let us ignore the requirement that the $a_n$ be natural numbers. Let $\varepsilon_m>0$ be a sequence of numbers tending very slowly to zero (e.g. $\varepsilon_m = 1 / \log \log(m+100)$). Define $a_n$ so that the ratios take the form $$ \frac{a_{4m+1}}{a_{4m}} = (1 + \varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+2}}{a_{4m+1}} = (1 + 2\varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+3}}{a_{4m+2}} = (1 + 3\varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+4}}{a_{4m+3}} = (1 + 2\varepsilon_m)^{1/2}$$ (setting $a_1=1$ say). Then $\sum_n (-1)^n (\frac{a_{n+1}}{a_n} - 1)^2$$\sum_n (-1)^n ((\frac{a_{n+1}}{a_n})^2 - 1)$ converges (indeed the series is designed so that the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ vanish, and the summands go to zero). On the other hand, from Taylor expansion, the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ of $\sum_n (-1)^n (\frac{a_{n+1}}{a_n} - 1)$ sums to $-\frac{1}{4} \varepsilon_m^2 + O(\varepsilon_m^3)$ if I did the arithmetic correctly, and this will diverge if $\varepsilon_m$ decays slowly enough.

To make the $a_n$ integer, one should take the real sequence constructed above and replace each element by its integer part. The sequence grows almost exponentially fast, so one can check that the errors incurred by doing so do not disrupt the convergence or divergence.

One can concoct similar examples for other $k$, or assuming the convergence or divergence of the other series mentioned in the post. More generally, once the summands are allowed to oscillate, the (conditional) convergence of one series says almost nothing about the convergence of any other series due to the lack of any useful comparison inequalities for oscillating sums.

No, one can create sequences in which the first sequence converges but the second does not (or vice versa).

For sake of argument take $k=2$. To begin with let us ignore the requirement that the $a_n$ be natural numbers. Let $\varepsilon_m>0$ be a sequence of numbers tending very slowly to zero (e.g. $\varepsilon_m = 1 / \log \log(m+100)$). Define $a_n$ so that the ratios take the form $$ \frac{a_{4m+1}}{a_{4m}} = (1 + \varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+2}}{a_{4m+1}} = (1 + 2\varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+3}}{a_{4m+2}} = (1 + 3\varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+4}}{a_{4m+3}} = (1 + 2\varepsilon_m)^{1/2}$$ (setting $a_1=1$ say). Then $\sum_n (-1)^n (\frac{a_{n+1}}{a_n} - 1)^2$ converges (indeed the series is designed so that the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ vanish, and the summands go to zero). On the other hand, from Taylor expansion, the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ of $\sum_n (-1)^n (\frac{a_{n+1}}{a_n} - 1)$ sums to $-\frac{1}{4} \varepsilon_m^2 + O(\varepsilon_m^3)$ if I did the arithmetic correctly, and this will diverge if $\varepsilon_m$ decays slowly enough.

To make the $a_n$ integer, one should take the real sequence constructed above and replace each element by its integer part. The sequence grows almost exponentially fast, so one can check that the errors incurred by doing so do not disrupt the convergence or divergence.

One can concoct similar examples for other $k$, or assuming the convergence or divergence of the other series mentioned in the post. More generally, once the summands are allowed to oscillate, the (conditional) convergence of one series says almost nothing about the convergence of any other series due to the lack of any useful comparison inequalities for oscillating sums.

No, one can create sequences in which the first sequence converges but the second does not (or vice versa).

For sake of argument take $k=2$. To begin with let us ignore the requirement that the $a_n$ be natural numbers. Let $\varepsilon_m>0$ be a sequence of numbers tending very slowly to zero (e.g. $\varepsilon_m = 1 / \log \log(m+100)$). Define $a_n$ so that the ratios take the form $$ \frac{a_{4m+1}}{a_{4m}} = (1 + \varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+2}}{a_{4m+1}} = (1 + 2\varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+3}}{a_{4m+2}} = (1 + 3\varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+4}}{a_{4m+3}} = (1 + 2\varepsilon_m)^{1/2}$$ (setting $a_1=1$ say). Then $\sum_n (-1)^n ((\frac{a_{n+1}}{a_n})^2 - 1)$ converges (indeed the series is designed so that the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ vanish, and the summands go to zero). On the other hand, from Taylor expansion, the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ of $\sum_n (-1)^n (\frac{a_{n+1}}{a_n} - 1)$ sums to $-\frac{1}{4} \varepsilon_m^2 + O(\varepsilon_m^3)$ if I did the arithmetic correctly, and this will diverge if $\varepsilon_m$ decays slowly enough.

To make the $a_n$ integer, one should take the real sequence constructed above and replace each element by its integer part. The sequence grows almost exponentially fast, so one can check that the errors incurred by doing so do not disrupt the convergence or divergence.

One can concoct similar examples for other $k$, or assuming the convergence or divergence of the other series mentioned in the post. More generally, once the summands are allowed to oscillate, the (conditional) convergence of one series says almost nothing about the convergence of any other series due to the lack of any useful comparison inequalities for oscillating sums.

Source Link
Terry Tao
Terry Tao
  • 114.1k
  • 33
  • 462
  • 539

No, one can create sequences in which the first sequence converges but the second does not (or vice versa).

For sake of argument take $k=2$. To begin with let us ignore the requirement that the $a_n$ be natural numbers. Let $\varepsilon_m>0$ be a sequence of numbers tending very slowly to zero (e.g. $\varepsilon_m = 1 / \log \log(m+100)$). Define $a_n$ so that the ratios take the form $$ \frac{a_{4m+1}}{a_{4m}} = (1 + \varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+2}}{a_{4m+1}} = (1 + 2\varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+3}}{a_{4m+2}} = (1 + 3\varepsilon_m)^{1/2} $$ $$ \frac{a_{4m+4}}{a_{4m+3}} = (1 + 2\varepsilon_m)^{1/2}$$ (setting $a_1=1$ say). Then $\sum_n (-1)^n (\frac{a_{n+1}}{a_n} - 1)^2$ converges (indeed the series is designed so that the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ vanish, and the summands go to zero). On the other hand, from Taylor expansion, the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ of $\sum_n (-1)^n (\frac{a_{n+1}}{a_n} - 1)$ sums to $-\frac{1}{4} \varepsilon_m^2 + O(\varepsilon_m^3)$ if I did the arithmetic correctly, and this will diverge if $\varepsilon_m$ decays slowly enough.

To make the $a_n$ integer, one should take the real sequence constructed above and replace each element by its integer part. The sequence grows almost exponentially fast, so one can check that the errors incurred by doing so do not disrupt the convergence or divergence.

One can concoct similar examples for other $k$, or assuming the convergence or divergence of the other series mentioned in the post. More generally, once the summands are allowed to oscillate, the (conditional) convergence of one series says almost nothing about the convergence of any other series due to the lack of any useful comparison inequalities for oscillating sums.