Mixing
Convergence to zero of a specific infinite product
0
Maybe less ambitious but I would be happy to prove the following. I have tried but without success.
Thanks for any help
infinite-product
asked Nov 20 '18 at 19:54
Gianfranco OLDANI
765
add a comment |
0
Maybe less ambitious but I would be happy to prove the following. I have tried but without success.
Thanks for any help
infinite-product
asked Nov 20 '18 at 19:54
Gianfranco OLDANI
765
After some work I can show that $ prod (1 - (1-p_n)^n)$ behaves exactly as $ prod (1 - e^{-np_n})$ . Now the answer to my previous question is a simple corollary. Now, giving the decreasing sequence $0 < p_n <= 1$ the next and probably more challenging question is to determine the right behaviour of the sequence $np_n$ in order to have the product $prod (1 - (1-p_n)^n)$ being 0 or positive. The conjecture is the following: Let $p_n = a log(n)/n$ with a real, $ 0 le a$ then the product is not nul for 1 < a and nul for 0<= a <=1 Counterexamples? Thank a lot
– Gianfranco OLDANI
Dec 2 '18 at 13:08
add a comment |
0
0
0
Maybe less ambitious but I would be happy to prove the following. I have tried but without success.
Thanks for any help
infinite-product
asked Nov 20 '18 at 19:54
Gianfranco OLDANI
765
Maybe less ambitious but I would be happy to prove the following. I have tried but without success.
Thanks for any help
infinite-product
infinite-product
asked Nov 20 '18 at 19:54
Gianfranco OLDANI
765
asked Nov 20 '18 at 19:54
Gianfranco OLDANI
765
asked Nov 20 '18 at 19:54
Gianfranco OLDANI
765
asked Nov 20 '18 at 19:54
Gianfranco OLDANI
765
asked Nov 20 '18 at 19:54
Gianfranco OLDANI
765
765
After some work I can show that $ prod (1 - (1-p_n)^n)$ behaves exactly as $ prod (1 - e^{-np_n})$ . Now the answer to my previous question is a simple corollary. Now, giving the decreasing sequence $0 < p_n <= 1$ the next and probably more challenging question is to determine the right behaviour of the sequence $np_n$ in order to have the product $prod (1 - (1-p_n)^n)$ being 0 or positive. The conjecture is the following: Let $p_n = a log(n)/n$ with a real, $ 0 le a$ then the product is not nul for 1 < a and nul for 0<= a <=1 Counterexamples? Thank a lot
– Gianfranco OLDANI
Dec 2 '18 at 13:08
add a comment |
After some work I can show that $ prod (1 - (1-p_n)^n)$ behaves exactly as $ prod (1 - e^{-np_n})$ . Now the answer to my previous question is a simple corollary. Now, giving the decreasing sequence $0 < p_n <= 1$ the next and probably more challenging question is to determine the right behaviour of the sequence $np_n$ in order to have the product $prod (1 - (1-p_n)^n)$ being 0 or positive. The conjecture is the following: Let $p_n = a log(n)/n$ with a real, $ 0 le a$ then the product is not nul for 1 < a and nul for 0<= a <=1 Counterexamples? Thank a lot
– Gianfranco OLDANI
Dec 2 '18 at 13:08
After some work I can show that $ prod (1 - (1-p_n)^n)$ behaves exactly as $ prod (1 - e^{-np_n})$ . Now the answer to my previous question is a simple corollary. Now, giving the decreasing sequence $0 < p_n <= 1$ the next and probably more challenging question is to determine the right behaviour of the sequence $np_n$ in order to have the product $prod (1 - (1-p_n)^n)$ being 0 or positive. The conjecture is the following: Let $p_n = a log(n)/n$ with a real, $ 0 le a$ then the product is not nul for 1 < a and nul for 0<= a <=1 Counterexamples? Thank a lot
– Gianfranco OLDANI
Dec 2 '18 at 13:08
After some work I can show that $ prod (1 - (1-p_n)^n)$ behaves exactly as $ prod (1 - e^{-np_n})$ . Now the answer to my previous question is a simple corollary. Now, giving the decreasing sequence $0 < p_n <= 1$ the next and probably more challenging question is to determine the right behaviour of the sequence $np_n$ in order to have the product $prod (1 - (1-p_n)^n)$ being 0 or positive. The conjecture is the following: Let $p_n = a log(n)/n$ with a real, $ 0 le a$ then the product is not nul for 1 < a and nul for 0<= a <=1 Counterexamples? Thank a lot
– Gianfranco OLDANI
Dec 2 '18 at 13:08
add a comment |
1 Answer
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Hint:
$$ lim_{n to infty} (1-1/n)^n = 1/e$$
What does this tell you about the limit of $(1-1/n)^{8n+4}$?
answered Nov 20 '18 at 19:59
Robert Israel
318k23208457
Ho! yes of course, thank you Robert. Then after a certain point my product behaves like an infinite power of (1-1/e) and then converging to 0. Looking further , I have done numerical computations that seems to show that also the following product is zero: $$prod (1 - (1-p_n)^{8n+4}) with p_n= 1/(nlog(n))$$ But if I am not wrong, the above argument cannot be applied.
– Gianfranco OLDANI
Nov 21 '18 at 14:28
In this case $1 - (1 - p_n)^{8n+4} to 0$ as $n to infty$, so of course the product goes to $0$.
– Robert Israel
Nov 21 '18 at 15:14
More generally I am interested for which sequences of $p_n$ the infinite product converges to zero and for which it converges to a positive value. To this end, using the Euler formula that you mentioned above for $e$ if I write $ p_n $ $ as $n p_n/n $ $ can I estimate the previous product with$ prod (1 - e^{-np_n}) $ in the sense that both converges simultaneously to zero or to a positive value?
– Gianfranco OLDANI
Nov 26 '18 at 20:34
add a comment |
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1 Answer
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0
Hint:
$$ lim_{n to infty} (1-1/n)^n = 1/e$$
What does this tell you about the limit of $(1-1/n)^{8n+4}$?
answered Nov 20 '18 at 19:59
Robert Israel
318k23208457
Ho! yes of course, thank you Robert. Then after a certain point my product behaves like an infinite power of (1-1/e) and then converging to 0. Looking further , I have done numerical computations that seems to show that also the following product is zero: $$prod (1 - (1-p_n)^{8n+4}) with p_n= 1/(nlog(n))$$ But if I am not wrong, the above argument cannot be applied.
– Gianfranco OLDANI
Nov 21 '18 at 14:28
In this case $1 - (1 - p_n)^{8n+4} to 0$ as $n to infty$, so of course the product goes to $0$.
– Robert Israel
Nov 21 '18 at 15:14
More generally I am interested for which sequences of $p_n$ the infinite product converges to zero and for which it converges to a positive value. To this end, using the Euler formula that you mentioned above for $e$ if I write $ p_n $ $ as $n p_n/n $ $ can I estimate the previous product with$ prod (1 - e^{-np_n}) $ in the sense that both converges simultaneously to zero or to a positive value?
– Gianfranco OLDANI
Nov 26 '18 at 20:34
add a comment |
0
Hint:
$$ lim_{n to infty} (1-1/n)^n = 1/e$$
What does this tell you about the limit of $(1-1/n)^{8n+4}$?
answered Nov 20 '18 at 19:59
Robert Israel
318k23208457
Ho! yes of course, thank you Robert. Then after a certain point my product behaves like an infinite power of (1-1/e) and then converging to 0. Looking further , I have done numerical computations that seems to show that also the following product is zero: $$prod (1 - (1-p_n)^{8n+4}) with p_n= 1/(nlog(n))$$ But if I am not wrong, the above argument cannot be applied.
– Gianfranco OLDANI
Nov 21 '18 at 14:28
In this case $1 - (1 - p_n)^{8n+4} to 0$ as $n to infty$, so of course the product goes to $0$.
– Robert Israel
Nov 21 '18 at 15:14
More generally I am interested for which sequences of $p_n$ the infinite product converges to zero and for which it converges to a positive value. To this end, using the Euler formula that you mentioned above for $e$ if I write $ p_n $ $ as $n p_n/n $ $ can I estimate the previous product with$ prod (1 - e^{-np_n}) $ in the sense that both converges simultaneously to zero or to a positive value?
– Gianfranco OLDANI
Nov 26 '18 at 20:34
add a comment |
0
0
0
Hint:
$$ lim_{n to infty} (1-1/n)^n = 1/e$$
What does this tell you about the limit of $(1-1/n)^{8n+4}$?
answered Nov 20 '18 at 19:59
Robert Israel
318k23208457
Hint:
$$ lim_{n to infty} (1-1/n)^n = 1/e$$
What does this tell you about the limit of $(1-1/n)^{8n+4}$?
answered Nov 20 '18 at 19:59
Robert Israel
318k23208457
answered Nov 20 '18 at 19:59
Robert Israel
318k23208457
answered Nov 20 '18 at 19:59
Robert Israel
318k23208457
answered Nov 20 '18 at 19:59
Robert Israel
318k23208457
318k23208457
Ho! yes of course, thank you Robert. Then after a certain point my product behaves like an infinite power of (1-1/e) and then converging to 0. Looking further , I have done numerical computations that seems to show that also the following product is zero: $$prod (1 - (1-p_n)^{8n+4}) with p_n= 1/(nlog(n))$$ But if I am not wrong, the above argument cannot be applied.
– Gianfranco OLDANI
Nov 21 '18 at 14:28
In this case $1 - (1 - p_n)^{8n+4} to 0$ as $n to infty$, so of course the product goes to $0$.
– Robert Israel
Nov 21 '18 at 15:14
More generally I am interested for which sequences of $p_n$ the infinite product converges to zero and for which it converges to a positive value. To this end, using the Euler formula that you mentioned above for $e$ if I write $ p_n $ $ as $n p_n/n $ $ can I estimate the previous product with$ prod (1 - e^{-np_n}) $ in the sense that both converges simultaneously to zero or to a positive value?
– Gianfranco OLDANI
Nov 26 '18 at 20:34
add a comment |
Ho! yes of course, thank you Robert. Then after a certain point my product behaves like an infinite power of (1-1/e) and then converging to 0. Looking further , I have done numerical computations that seems to show that also the following product is zero: $$prod (1 - (1-p_n)^{8n+4}) with p_n= 1/(nlog(n))$$ But if I am not wrong, the above argument cannot be applied.
– Gianfranco OLDANI
Nov 21 '18 at 14:28
In this case $1 - (1 - p_n)^{8n+4} to 0$ as $n to infty$, so of course the product goes to $0$.
– Robert Israel
Nov 21 '18 at 15:14
More generally I am interested for which sequences of $p_n$ the infinite product converges to zero and for which it converges to a positive value. To this end, using the Euler formula that you mentioned above for $e$ if I write $ p_n $ $ as $n p_n/n $ $ can I estimate the previous product with$ prod (1 - e^{-np_n}) $ in the sense that both converges simultaneously to zero or to a positive value?
– Gianfranco OLDANI
Nov 26 '18 at 20:34
Ho! yes of course, thank you Robert. Then after a certain point my product behaves like an infinite power of (1-1/e) and then converging to 0. Looking further , I have done numerical computations that seems to show that also the following product is zero: $$prod (1 - (1-p_n)^{8n+4}) with p_n= 1/(nlog(n))$$ But if I am not wrong, the above argument cannot be applied.
– Gianfranco OLDANI
Nov 21 '18 at 14:28
Ho! yes of course, thank you Robert. Then after a certain point my product behaves like an infinite power of (1-1/e) and then converging to 0. Looking further , I have done numerical computations that seems to show that also the following product is zero: $$prod (1 - (1-p_n)^{8n+4}) with p_n= 1/(nlog(n))$$ But if I am not wrong, the above argument cannot be applied.
– Gianfranco OLDANI
Nov 21 '18 at 14:28
In this case $1 - (1 - p_n)^{8n+4} to 0$ as $n to infty$, so of course the product goes to $0$.
– Robert Israel
Nov 21 '18 at 15:14
In this case $1 - (1 - p_n)^{8n+4} to 0$ as $n to infty$, so of course the product goes to $0$.
– Robert Israel
Nov 21 '18 at 15:14
More generally I am interested for which sequences of $p_n$ the infinite product converges to zero and for which it converges to a positive value. To this end, using the Euler formula that you mentioned above for $e$ if I write $ p_n $ $ as $n p_n/n $ $ can I estimate the previous product with$ prod (1 - e^{-np_n}) $ in the sense that both converges simultaneously to zero or to a positive value?
– Gianfranco OLDANI
Nov 26 '18 at 20:34
More generally I am interested for which sequences of $p_n$ the infinite product converges to zero and for which it converges to a positive value. To this end, using the Euler formula that you mentioned above for $e$ if I write $ p_n $ $ as $n p_n/n $ $ can I estimate the previous product with$ prod (1 - e^{-np_n}) $ in the sense that both converges simultaneously to zero or to a positive value?
– Gianfranco OLDANI
Nov 26 '18 at 20:34
add a comment |
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After some work I can show that $ prod (1 - (1-p_n)^n)$ behaves exactly as $ prod (1 - e^{-np_n})$ . Now the answer to my previous question is a simple corollary. Now, giving the decreasing sequence $0 < p_n <= 1$ the next and probably more challenging question is to determine the right behaviour of the sequence $np_n$ in order to have the product $prod (1 - (1-p_n)^n)$ being 0 or positive. The conjecture is the following: Let $p_n = a log(n)/n$ with a real, $ 0 le a$ then the product is not nul for 1 < a and nul for 0<= a <=1 Counterexamples? Thank a lot
– Gianfranco OLDANI
Dec 2 '18 at 13:08