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Question Number 5948 by 123456 last updated on 06/Jun/16
is there a_n  such that Σa_n  and Πa_n   converge?
$$\mathrm{is}\:\mathrm{there}\:{a}_{{n}} \:\mathrm{such}\:\mathrm{that}\:\Sigma{a}_{{n}} \:\mathrm{and}\:\Pi{a}_{{n}} \\ $$$$\mathrm{converge}? \\ $$
Answered by Yozzii last updated on 06/Jun/16
Yes. Let a_n =r^n  where  ∣r∣<1.  ⇒Σ_(n=1) ^∞ r^n =(r/(1−r)) and Π_(n=1) ^∞ r^n =lim_(N→∞) Π_(n=1) ^N r^n =lim_(N→∞) r^(0.5N(N+1)) =r^∞ =0.  So, ∃a_n ∈R such that Σa_n  and Πa_n   both comverge. More interesting  examples could be found.
$${Yes}.\:{Let}\:{a}_{{n}} ={r}^{{n}} \:{where}\:\:\mid{r}\mid<\mathrm{1}. \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{r}^{{n}} =\frac{{r}}{\mathrm{1}−{r}}\:{and}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}{r}^{{n}} =\underset{{N}\rightarrow\infty} {\mathrm{lim}}\underset{{n}=\mathrm{1}} {\overset{{N}} {\prod}}{r}^{{n}} =\underset{{N}\rightarrow\infty} {\mathrm{lim}}{r}^{\mathrm{0}.\mathrm{5}{N}\left({N}+\mathrm{1}\right)} ={r}^{\infty} =\mathrm{0}. \\ $$$${So},\:\exists{a}_{{n}} \in\mathbb{R}\:{such}\:{that}\:\Sigma{a}_{{n}} \:{and}\:\Pi{a}_{{n}} \\ $$$${both}\:{comverge}.\:{More}\:{interesting} \\ $$$${examples}\:{could}\:{be}\:{found}. \\ $$

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