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Show that there is a sequence $\{t_n\}$, such that both $\sum_{n=1}^{\infty}t_n$ and $\Pi_{n=1}^{\infty} 1-2e^{-2(t_n)^2}$ converges.

Is it possible to make $\Pi_{n=1}^{\infty} 1-2e^{-2(t_n)^2}>0$ (but not necessarily convergent)?

My attempt: I've been thinking about this for a while. I think the most obvious choice of $t_n$ would be something like this: $\sqrt{ln(\dfrac{p_n-1}{-2})/(-2)}$. Then it is obvious that $\Pi_{n=1}^{\infty} 1-2e^{-2(t_n)^2}=\Pi_{n=1}^{\infty}p_n$, which means we only have to look for $p_n$ such that $\sum_{n=1}^{\infty} ln(p_n)$ converges. Then we can replace $q_n=ln(p_n)$.

The problem now simplifies to finding a $\{q_n\}$ such that $\sum_{n=1}^{\infty} q_n$ and $\sum_{n=1}^{\infty} \sqrt{ln(\dfrac{p_n-1}{-2})/(-2)}$ both converges. Then I don't know what to do next.

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  • $\begingroup$ You are certain it's $(1-2e^{-2t_n^2})$ in the product, and not $(2e^{-2t_n^2}-1)$? $\endgroup$ – Clement C. Sep 9 '17 at 21:40
  • $\begingroup$ @ClementC. I'm sure it is one minus instead of other way around. But why should it be the other way around? $\endgroup$ – Daniel Li Sep 9 '17 at 21:48
  • $\begingroup$ @ClementC. Why retract? Weren't you thinking that if the sum is to converge then necessarily $|t_n| \to 0$, then we must have each $1 - 2e^{2(t_n)^2} \to -1$, which seems flipped? $\endgroup$ – Lee David Chung Lin Sep 9 '17 at 22:12
  • $\begingroup$ I removed my comment because I was hoping to subsum it in an answer. As it turns out, I have to run now, so it's not going to happen now. First suggestion: try $t_n = (-1)^n/\sqrt n$? $\endgroup$ – Clement C. Sep 9 '17 at 22:17
  • $\begingroup$ The series will converges by the alternating series criterion. The product should go to zero since t_n goes to 0 too slowly. $\endgroup$ – Clement C. Sep 9 '17 at 22:21
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If $\sum_{n=1}^\infty t_n$ converges, $t_n \rightarrow 0$. But $$ \lim_{x \rightarrow 0} 1 - 2 \mathrm{e}^{-2 x^2} = -1 \text{,} $$ so your product cannot converge by having terms $\rightarrow 1$. Maybe we can get this to work the other way around, having the product "diverge to $0$"...

Suppose we want, for $n > N \in \mathbb{N}$, $-1 < 1 - 2 \mathrm{e}^{-2 t_n^2} < -1 + \epsilon$ so that (eventually) all the terms are less than unit magnitude and we can arrange for them to converge to $-1$ so that the sum may converge. Then $$ 0 < t_n < \frac{1}{\sqrt{2}} \sqrt{-\ln \frac{2-\epsilon}{2}} \text{.} $$ If we try $\epsilon \mapsto n^{-n}$, $t_n = \frac{1}{\sqrt{2}} \sqrt{-\ln \frac{2-n^{-n}}{2}}$, and then the product is $\prod_{n=1}^\infty -1+n^{-n} = 0$, which converges by diverging to $0$.

Now the logarithm is concave down everywhere and passes through $(1,0)$, so \begin{align*} \frac{1}{\sqrt{2}} \sqrt{-\ln \frac{2-n^{-n}}{2}} &= \frac{1}{\sqrt{2}} \sqrt{\ln \frac{2}{2-n^{-n}}} \\ &= \frac{1}{\sqrt{2}} \sqrt{\ln ( 1 + \frac{n^{-n}}{2-n^{-n}}}) \\ &< \frac{1}{\sqrt{2}} \sqrt{\frac{n^{-n}}{2-n^{-n}}} \\ &< \frac{1}{\sqrt{2}} \sqrt{\frac{n^{-n}}{1}} \\ &= \frac{1}{\sqrt{2}} n^{-n/2} \text{.} \end{align*} Then, since $\sum_{n=1}^\infty n^{-n/2}$ converges, so does $\sum_{i=1}^n t_n$.

In retrospect, $\epsilon \mapsto 2n^{-2n}$ might have been a little nicer to push through the comparison test.

As I noted in a comment $$ t_n = \begin{cases} -2, &n = 1 \\ 3^{1-n}, &n > 1 \end{cases} $$ makes your product $-3 \cdot \prod_{n=1}^\infty 1 - 2 \mathrm{e}^{-2 (3^{-n})^2} = -3 \cdot -(1/3; 1/3)_\infty = 1.68038\dots$ and your sum still converges (since a finite initial sequence has no impact on convergence). We can make the product converge to whatever value you like by altering the one inserted sequence member.

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  • $\begingroup$ Is it possible to make $\Pi_{n=1}^{\infty} 1-2e^{-2(t_n)^2}>0$ (but not necessarily convergent)? $\endgroup$ – Daniel Li Sep 9 '17 at 22:52
  • $\begingroup$ @DanielLi : I don't understand how you're ascribing a value to an infinite product without requiring it to converge. $\endgroup$ – Eric Towers Sep 9 '17 at 23:01
  • $\begingroup$ @DanielLi : If we set $t_n \mapsto 3^{-n}$, then the product converges to $(1/3; 1/3)_\infty = -0.560126\dots$, using the q-Pochhammer symbol. If we prepend $-2$ to the sequence, then the prepended multiplicand is $-3$, which makes the resulting product positive. Convergence of the sum is indifferent to prepending this term to the sequence. $\endgroup$ – Eric Towers Sep 9 '17 at 23:12
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Late to the party, but elaborating on my comment above:

Choose $(t_n)_n$ defined by $$t_n = \frac{(-1)^n}{\sqrt{n}},\quad n\geq 1 \tag{1}$$ Then

  • the series s$\sum_{n=1}^\infty t_n$ converges (conditionally) by the Leibniz criterion.

  • the product $\prod_{n=1}^\infty (1-2e^{-2t_n^2})$ converges, to $0$: $$ \left\lvert \prod_{n=1}^N (1-2e^{-2t_n^2}) \right\rvert = \prod_{n=1}^N \left\lvert1-2e^{-2t_n^2} \right\rvert = \prod_{n=1}^N \left\lvert1-2e^{-2/n} \right\rvert = \exp \sum_{n=1}^N \ln\left(2e^{-2/n} -1\right) $$ and since $$\begin{align} \ln\left(2e^{-2/n} -1\right) &= \ln\left(2\left(1-\frac{2}{n}+o\left(\frac{1}{n}\right) \right)-1\right) = \ln\left(1-\frac{4}{n}+o\left(\frac{1}{n}\right) \right)\\ &= -\frac{4}{n}+o\left(\frac{1}{n}\right) \end{align}$$ by comparison with the Harmonic series we have $$\sum_{n=1}^N \ln\left(2e^{-2/n} -1\right)\operatorname*{\sim}_{N\to\infty} -4\sum_{n=1}^N \frac{1}{n} \xrightarrow[N\to\infty]{} -\infty$$ so that $$ \left\lvert \prod_{n=1}^N (1-2e^{-2t_n^2}) \right\rvert = \exp \sum_{n=1}^N \ln\left(2e^{-2/n} -1\right)\xrightarrow[N\to\infty]{} 0 \tag{2} $$ as wanted.

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