Convergence value of series with exponentialsProving the convergence of a productAbsolute and conditional convergence of function seriesTesting Convergence With Limit Comparison Testwhat value series $sum_r=1^inftyfraclfloor rp rfloor2^r$ converges?Open question about limits of functionsseries convergence with comparison test $frac1nlog(n)^p$Relation of the expectation value of the boltzmann distribution and the “temperature”Convergence of series: hintHow to find the converging value of the following voltage series (or at least the general term)?Is the series $sum_n=1^infty Bigl(1-Bigl(1-frac1n^1+epsilonBigr)^nBigr)$ convergent?
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Convergence value of series with exponentials
Proving the convergence of a productAbsolute and conditional convergence of function seriesTesting Convergence With Limit Comparison Testwhat value series $sum_r=1^inftyfraclfloor rp rfloor2^r$ converges?Open question about limits of functionsseries convergence with comparison test $frac1nlog(n)^p$Relation of the expectation value of the boltzmann distribution and the “temperature”Convergence of series: hintHow to find the converging value of the following voltage series (or at least the general term)?Is the series $sum_n=1^infty Bigl(1-Bigl(1-frac1n^1+epsilonBigr)^nBigr)$ convergent?
$begingroup$
I was solving a problem and I came across the following expression,
$$sum_n^N N choose nexp[-beta nomega]$$
I was looking for the convergence of this series but I couldn't find any resources which relates this sum of exponentials with a combinatory factor, so I was wondering if you have seen this series before or know any table where I can find general series for converging exponentials.
sequences-and-series exponential-sum
edited Mar 20 at 18:58
Bernard
124k741118
asked Mar 20 at 18:17
CharlieCharlie
1496
$endgroup$
add a comment |
$begingroup$
I was solving a problem and I came across the following expression,
$$sum_n^N N choose nexp[-beta nomega]$$
I was looking for the convergence of this series but I couldn't find any resources which relates this sum of exponentials with a combinatory factor, so I was wondering if you have seen this series before or know any table where I can find general series for converging exponentials.
sequences-and-series exponential-sum
edited Mar 20 at 18:58
Bernard
124k741118
asked Mar 20 at 18:17
CharlieCharlie
1496
$endgroup$
$begingroup$
Let $x=exp[-betaomega]$. You series is $sum_nbinomNnx^n$. Does this look familiar?
$endgroup$
– Mike Earnest
Mar 20 at 18:25
$begingroup$
Sum on $i$ or sum on $n$ ?
$endgroup$
– TheSilverDoe
Mar 20 at 18:28
$begingroup$
It seems to be a finite sum; how could convergence be a question?
$endgroup$
– Greg Martin
Mar 20 at 18:28
$begingroup$
@MikeEarnest Indeed, it remembers me to the binomial theorem with $sum_n N choose n x^n=(1+x)^n$, so substituting with $x=exp[-beta omega]$ I get $(1+exp[-betaomega])^n$.
$endgroup$
– Charlie
Mar 20 at 18:33
add a comment |
$begingroup$
I was solving a problem and I came across the following expression,
$$sum_n^N N choose nexp[-beta nomega]$$
I was looking for the convergence of this series but I couldn't find any resources which relates this sum of exponentials with a combinatory factor, so I was wondering if you have seen this series before or know any table where I can find general series for converging exponentials.
sequences-and-series exponential-sum
edited Mar 20 at 18:58
Bernard
124k741118
asked Mar 20 at 18:17
CharlieCharlie
1496
$endgroup$
I was solving a problem and I came across the following expression,
$$sum_n^N N choose nexp[-beta nomega]$$
I was looking for the convergence of this series but I couldn't find any resources which relates this sum of exponentials with a combinatory factor, so I was wondering if you have seen this series before or know any table where I can find general series for converging exponentials.
sequences-and-series exponential-sum
sequences-and-series exponential-sum
edited Mar 20 at 18:58
Bernard
124k741118
asked Mar 20 at 18:17
CharlieCharlie
1496
edited Mar 20 at 18:58
Bernard
124k741118
asked Mar 20 at 18:17
CharlieCharlie
1496
edited Mar 20 at 18:58
Bernard
124k741118
edited Mar 20 at 18:58
Bernard
124k741118
edited Mar 20 at 18:58
Bernard
124k741118
124k741118
asked Mar 20 at 18:17
CharlieCharlie
1496
asked Mar 20 at 18:17
CharlieCharlie
1496
asked Mar 20 at 18:17
CharlieCharlie
1496
1496
$begingroup$
Let $x=exp[-betaomega]$. You series is $sum_nbinomNnx^n$. Does this look familiar?
$endgroup$
– Mike Earnest
Mar 20 at 18:25
$begingroup$
Sum on $i$ or sum on $n$ ?
$endgroup$
– TheSilverDoe
Mar 20 at 18:28
$begingroup$
It seems to be a finite sum; how could convergence be a question?
$endgroup$
– Greg Martin
Mar 20 at 18:28
$begingroup$
@MikeEarnest Indeed, it remembers me to the binomial theorem with $sum_n N choose n x^n=(1+x)^n$, so substituting with $x=exp[-beta omega]$ I get $(1+exp[-betaomega])^n$.
$endgroup$
– Charlie
Mar 20 at 18:33
add a comment |
$begingroup$
Let $x=exp[-betaomega]$. You series is $sum_nbinomNnx^n$. Does this look familiar?
$endgroup$
– Mike Earnest
Mar 20 at 18:25
$begingroup$
Sum on $i$ or sum on $n$ ?
$endgroup$
– TheSilverDoe
Mar 20 at 18:28
$begingroup$
It seems to be a finite sum; how could convergence be a question?
$endgroup$
– Greg Martin
Mar 20 at 18:28
$begingroup$
@MikeEarnest Indeed, it remembers me to the binomial theorem with $sum_n N choose n x^n=(1+x)^n$, so substituting with $x=exp[-beta omega]$ I get $(1+exp[-betaomega])^n$.
$endgroup$
– Charlie
Mar 20 at 18:33
$begingroup$
Let $x=exp[-betaomega]$. You series is $sum_nbinomNnx^n$. Does this look familiar?
$endgroup$
– Mike Earnest
Mar 20 at 18:25
$begingroup$
Let $x=exp[-betaomega]$. You series is $sum_nbinomNnx^n$. Does this look familiar?
$endgroup$
– Mike Earnest
Mar 20 at 18:25
$begingroup$
Sum on $i$ or sum on $n$ ?
$endgroup$
– TheSilverDoe
Mar 20 at 18:28
$begingroup$
Sum on $i$ or sum on $n$ ?
$endgroup$
– TheSilverDoe
Mar 20 at 18:28
$begingroup$
It seems to be a finite sum; how could convergence be a question?
$endgroup$
– Greg Martin
Mar 20 at 18:28
$begingroup$
It seems to be a finite sum; how could convergence be a question?
$endgroup$
– Greg Martin
Mar 20 at 18:28
$begingroup$
@MikeEarnest Indeed, it remembers me to the binomial theorem with $sum_n N choose n x^n=(1+x)^n$, so substituting with $x=exp[-beta omega]$ I get $(1+exp[-betaomega])^n$.
$endgroup$
– Charlie
Mar 20 at 18:33
$begingroup$
@MikeEarnest Indeed, it remembers me to the binomial theorem with $sum_n N choose n x^n=(1+x)^n$, so substituting with $x=exp[-beta omega]$ I get $(1+exp[-betaomega])^n$.
$endgroup$
– Charlie
Mar 20 at 18:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This sum is of the form
$$
sum_n binomNn a^n
$$
where $a = e^-betaomega$.
So by the binomial theorem, this is equal to
$$
sum_n binomNn a^n = (1+a)^n
$$
and the answer is
$$
(1+e^-betaomega
)^n
$$
answered Mar 20 at 18:30
Mark FischlerMark Fischler
33.9k12552
$endgroup$
$begingroup$
Thank you, I did what they suggested above and I got the same result.
$endgroup$
– Charlie
Mar 20 at 18:34
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This sum is of the form
$$
sum_n binomNn a^n
$$
where $a = e^-betaomega$.
So by the binomial theorem, this is equal to
$$
sum_n binomNn a^n = (1+a)^n
$$
and the answer is
$$
(1+e^-betaomega
)^n
$$
answered Mar 20 at 18:30
Mark FischlerMark Fischler
33.9k12552
$endgroup$
$begingroup$
Thank you, I did what they suggested above and I got the same result.
$endgroup$
– Charlie
Mar 20 at 18:34
add a comment |
$begingroup$
This sum is of the form
$$
sum_n binomNn a^n
$$
where $a = e^-betaomega$.
So by the binomial theorem, this is equal to
$$
sum_n binomNn a^n = (1+a)^n
$$
and the answer is
$$
(1+e^-betaomega
)^n
$$
answered Mar 20 at 18:30
Mark FischlerMark Fischler
33.9k12552
$endgroup$
$begingroup$
Thank you, I did what they suggested above and I got the same result.
$endgroup$
– Charlie
Mar 20 at 18:34
add a comment |
$begingroup$
This sum is of the form
$$
sum_n binomNn a^n
$$
where $a = e^-betaomega$.
So by the binomial theorem, this is equal to
$$
sum_n binomNn a^n = (1+a)^n
$$
and the answer is
$$
(1+e^-betaomega
)^n
$$
answered Mar 20 at 18:30
Mark FischlerMark Fischler
33.9k12552
$endgroup$
This sum is of the form
$$
sum_n binomNn a^n
$$
where $a = e^-betaomega$.
So by the binomial theorem, this is equal to
$$
sum_n binomNn a^n = (1+a)^n
$$
and the answer is
$$
(1+e^-betaomega
)^n
$$
answered Mar 20 at 18:30
Mark FischlerMark Fischler
33.9k12552
answered Mar 20 at 18:30
Mark FischlerMark Fischler
33.9k12552
answered Mar 20 at 18:30
Mark FischlerMark Fischler
33.9k12552
answered Mar 20 at 18:30
Mark FischlerMark Fischler
33.9k12552
33.9k12552
$begingroup$
Thank you, I did what they suggested above and I got the same result.
$endgroup$
– Charlie
Mar 20 at 18:34
add a comment |
$begingroup$
Thank you, I did what they suggested above and I got the same result.
$endgroup$
– Charlie
Mar 20 at 18:34
$begingroup$
Thank you, I did what they suggested above and I got the same result.
$endgroup$
– Charlie
Mar 20 at 18:34
$begingroup$
Thank you, I did what they suggested above and I got the same result.
$endgroup$
– Charlie
Mar 20 at 18:34
add a comment |
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$begingroup$
Let $x=exp[-betaomega]$. You series is $sum_nbinomNnx^n$. Does this look familiar?
$endgroup$
– Mike Earnest
Mar 20 at 18:25
$begingroup$
Sum on $i$ or sum on $n$ ?
$endgroup$
– TheSilverDoe
Mar 20 at 18:28
$begingroup$
It seems to be a finite sum; how could convergence be a question?
$endgroup$
– Greg Martin
Mar 20 at 18:28
$begingroup$
@MikeEarnest Indeed, it remembers me to the binomial theorem with $sum_n N choose n x^n=(1+x)^n$, so substituting with $x=exp[-beta omega]$ I get $(1+exp[-betaomega])^n$.
$endgroup$
– Charlie
Mar 20 at 18:33