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Question about convergence of infinite products
Infinite products (involving complex numbers)Convergence of a productHelp with Convergence/DivergenceA basic sequence convergence questionQuestion about series increasing and convergenceProof about the convergence of a sequenceStatements regarding sequences and seriesInfinite series: Integral test( Proof )Question regarding implication of series convergence.Convergence of the ratio of two infinite series.
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Because $lim_nrightarrow infty(1+fracxn)^n = e^x$, is it true that $prod_n(1+x_n)$ is bounded above by $e^sum_n=1^infty x_n$ and hence $prod_n(1+x_n)$ converges or diverges depending on $sum_n=1^inftyx_n$? Thanks!
calculus sequences-and-series
asked Mar 14 at 16:18
manifoldedmanifolded
48819
$endgroup$
add a comment |
$begingroup$
Because $lim_nrightarrow infty(1+fracxn)^n = e^x$, is it true that $prod_n(1+x_n)$ is bounded above by $e^sum_n=1^infty x_n$ and hence $prod_n(1+x_n)$ converges or diverges depending on $sum_n=1^inftyx_n$? Thanks!
calculus sequences-and-series
asked Mar 14 at 16:18
manifoldedmanifolded
48819
$endgroup$
$begingroup$
What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
$endgroup$
– Yves Daoust
Mar 14 at 16:52
$begingroup$
$prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
$endgroup$
– manifolded
Mar 14 at 17:06
$begingroup$
"bounded above" makes it a very different question !
$endgroup$
– Yves Daoust
Mar 14 at 17:10
$begingroup$
True, my bad, don't know why I did that. Thanks for the answer.
$endgroup$
– manifolded
Mar 14 at 17:11
add a comment |
$begingroup$
Because $lim_nrightarrow infty(1+fracxn)^n = e^x$, is it true that $prod_n(1+x_n)$ is bounded above by $e^sum_n=1^infty x_n$ and hence $prod_n(1+x_n)$ converges or diverges depending on $sum_n=1^inftyx_n$? Thanks!
calculus sequences-and-series
asked Mar 14 at 16:18
manifoldedmanifolded
48819
$endgroup$
Because $lim_nrightarrow infty(1+fracxn)^n = e^x$, is it true that $prod_n(1+x_n)$ is bounded above by $e^sum_n=1^infty x_n$ and hence $prod_n(1+x_n)$ converges or diverges depending on $sum_n=1^inftyx_n$? Thanks!
calculus sequences-and-series
calculus sequences-and-series
asked Mar 14 at 16:18
manifoldedmanifolded
48819
asked Mar 14 at 16:18
manifoldedmanifolded
48819
edited Mar 14 at 17:08
manifolded
asked Mar 14 at 16:18
manifoldedmanifolded
48819
asked Mar 14 at 16:18
manifoldedmanifolded
48819
asked Mar 14 at 16:18
manifoldedmanifolded
48819
48819
$begingroup$
What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
$endgroup$
– Yves Daoust
Mar 14 at 16:52
$begingroup$
$prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
$endgroup$
– manifolded
Mar 14 at 17:06
$begingroup$
"bounded above" makes it a very different question !
$endgroup$
– Yves Daoust
Mar 14 at 17:10
$begingroup$
True, my bad, don't know why I did that. Thanks for the answer.
$endgroup$
– manifolded
Mar 14 at 17:11
add a comment |
$begingroup$
What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
$endgroup$
– Yves Daoust
Mar 14 at 16:52
$begingroup$
$prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
$endgroup$
– manifolded
Mar 14 at 17:06
$begingroup$
"bounded above" makes it a very different question !
$endgroup$
– Yves Daoust
Mar 14 at 17:10
$begingroup$
True, my bad, don't know why I did that. Thanks for the answer.
$endgroup$
– manifolded
Mar 14 at 17:11
$begingroup$
What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
$endgroup$
– Yves Daoust
Mar 14 at 16:52
$begingroup$
What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
$endgroup$
– Yves Daoust
Mar 14 at 16:52
$begingroup$
$prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
$endgroup$
– manifolded
Mar 14 at 17:06
$begingroup$
$prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
$endgroup$
– manifolded
Mar 14 at 17:06
$begingroup$
"bounded above" makes it a very different question !
$endgroup$
– Yves Daoust
Mar 14 at 17:10
$begingroup$
"bounded above" makes it a very different question !
$endgroup$
– Yves Daoust
Mar 14 at 17:10
$begingroup$
True, my bad, don't know why I did that. Thanks for the answer.
$endgroup$
– manifolded
Mar 14 at 17:11
$begingroup$
True, my bad, don't know why I did that. Thanks for the answer.
$endgroup$
– manifolded
Mar 14 at 17:11
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$
and there is no reason that
$$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$
Given the change of the question, this answer is irrelevant.
answered Mar 14 at 17:09
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$
and there is no reason that
$$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$
Given the change of the question, this answer is irrelevant.
answered Mar 14 at 17:09
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$
and there is no reason that
$$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$
Given the change of the question, this answer is irrelevant.
answered Mar 14 at 17:09
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$
and there is no reason that
$$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$
Given the change of the question, this answer is irrelevant.
answered Mar 14 at 17:09
Yves DaoustYves Daoust
131k676229
$endgroup$
If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$
and there is no reason that
$$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$
Given the change of the question, this answer is irrelevant.
answered Mar 14 at 17:09
Yves DaoustYves Daoust
131k676229
answered Mar 14 at 17:09
Yves DaoustYves Daoust
131k676229
answered Mar 14 at 17:09
Yves DaoustYves Daoust
131k676229
answered Mar 14 at 17:09
Yves DaoustYves Daoust
131k676229
131k676229
add a comment |
add a comment |
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$begingroup$
What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
$endgroup$
– Yves Daoust
Mar 14 at 16:52
$begingroup$
$prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
$endgroup$
– manifolded
Mar 14 at 17:06
$begingroup$
"bounded above" makes it a very different question !
$endgroup$
– Yves Daoust
Mar 14 at 17:10
$begingroup$
True, my bad, don't know why I did that. Thanks for the answer.
$endgroup$
– manifolded
Mar 14 at 17:11