Computation of a limit involving a series (related to Poisson distribution)Complete statistic: Poisson DistributionUnbiased estimator - Poisson DistributionLimit involving $sqrt[n]n!$Proving that if $a_ngeq0$ and $sum a_n$ converges, then $sum a_n^2$ convergesHelp me out by calculating this limitLimit of series/n convergenceImplications of Poisson distributionLimit of the series $lim_nrightarrow inftyfrac1s_nsum_k=1^na_kx_k$Not understanding a proof for the formula for Poisson distributionDoes this limit exist on $mathbb R^2$
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Computation of a limit involving a series (related to Poisson distribution)
Complete statistic: Poisson DistributionUnbiased estimator - Poisson DistributionLimit involving $sqrt[n]n!$Proving that if $a_ngeq0$ and $sum a_n$ converges, then $sum a_n^2$ convergesHelp me out by calculating this limitLimit of series/n convergenceImplications of Poisson distributionLimit of the series $lim_nrightarrow inftyfrac1s_nsum_k=1^na_kx_k$Not understanding a proof for the formula for Poisson distributionDoes this limit exist on $mathbb R^2$
$begingroup$
Consider $lambda >0.$ I am reading a paper and the author states that
$$ displaystylelim_v rightarrow +infty sum_n=0^+infty fraclambda^n(n !)^v = 1 + lambda$$
I tried to compute such limit but I am getting anywhere. Someone could help me?
Thanks in advance!
analysis statistics
asked 15 hours ago
math studentmath student
2,39111018
$endgroup$
add a comment |
$begingroup$
Consider $lambda >0.$ I am reading a paper and the author states that
$$ displaystylelim_v rightarrow +infty sum_n=0^+infty fraclambda^n(n !)^v = 1 + lambda$$
I tried to compute such limit but I am getting anywhere. Someone could help me?
Thanks in advance!
analysis statistics
asked 15 hours ago
math studentmath student
2,39111018
$endgroup$
add a comment |
$begingroup$
Consider $lambda >0.$ I am reading a paper and the author states that
$$ displaystylelim_v rightarrow +infty sum_n=0^+infty fraclambda^n(n !)^v = 1 + lambda$$
I tried to compute such limit but I am getting anywhere. Someone could help me?
Thanks in advance!
analysis statistics
asked 15 hours ago
math studentmath student
2,39111018
$endgroup$
Consider $lambda >0.$ I am reading a paper and the author states that
$$ displaystylelim_v rightarrow +infty sum_n=0^+infty fraclambda^n(n !)^v = 1 + lambda$$
I tried to compute such limit but I am getting anywhere. Someone could help me?
Thanks in advance!
analysis statistics
analysis statistics
asked 15 hours ago
math studentmath student
2,39111018
asked 15 hours ago
math studentmath student
2,39111018
asked 15 hours ago
math studentmath student
2,39111018
asked 15 hours ago
math studentmath student
2,39111018
asked 15 hours ago
math studentmath student
2,39111018
2,39111018
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Note that $$sum_n=0^+infty fraclambda^n(n !)^v=1+lambda+sum_n=2^+infty fraclambda^n(n !)^v$$and $$sum_n=2^+infty fraclambda^n(n !)^vle sum_n=2^+infty fraclambda^nn !sum_n=2^+infty frac1(n !)^v-1\le e^lambdasum_n=2^infty1over 2^(n-1)(v-1)\=e^lambda1over 2^v-1over 1-1over 2^v-1\=e^lambdaover 2^v-1-1\to 0$$hence the result.
answered 15 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
$endgroup$
add a comment |
$begingroup$
The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).
answered 15 hours ago
user66081user66081
3,2581126
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that $$sum_n=0^+infty fraclambda^n(n !)^v=1+lambda+sum_n=2^+infty fraclambda^n(n !)^v$$and $$sum_n=2^+infty fraclambda^n(n !)^vle sum_n=2^+infty fraclambda^nn !sum_n=2^+infty frac1(n !)^v-1\le e^lambdasum_n=2^infty1over 2^(n-1)(v-1)\=e^lambda1over 2^v-1over 1-1over 2^v-1\=e^lambdaover 2^v-1-1\to 0$$hence the result.
answered 15 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
$endgroup$
add a comment |
$begingroup$
Note that $$sum_n=0^+infty fraclambda^n(n !)^v=1+lambda+sum_n=2^+infty fraclambda^n(n !)^v$$and $$sum_n=2^+infty fraclambda^n(n !)^vle sum_n=2^+infty fraclambda^nn !sum_n=2^+infty frac1(n !)^v-1\le e^lambdasum_n=2^infty1over 2^(n-1)(v-1)\=e^lambda1over 2^v-1over 1-1over 2^v-1\=e^lambdaover 2^v-1-1\to 0$$hence the result.
answered 15 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
$endgroup$
add a comment |
$begingroup$
Note that $$sum_n=0^+infty fraclambda^n(n !)^v=1+lambda+sum_n=2^+infty fraclambda^n(n !)^v$$and $$sum_n=2^+infty fraclambda^n(n !)^vle sum_n=2^+infty fraclambda^nn !sum_n=2^+infty frac1(n !)^v-1\le e^lambdasum_n=2^infty1over 2^(n-1)(v-1)\=e^lambda1over 2^v-1over 1-1over 2^v-1\=e^lambdaover 2^v-1-1\to 0$$hence the result.
answered 15 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
$endgroup$
Note that $$sum_n=0^+infty fraclambda^n(n !)^v=1+lambda+sum_n=2^+infty fraclambda^n(n !)^v$$and $$sum_n=2^+infty fraclambda^n(n !)^vle sum_n=2^+infty fraclambda^nn !sum_n=2^+infty frac1(n !)^v-1\le e^lambdasum_n=2^infty1over 2^(n-1)(v-1)\=e^lambda1over 2^v-1over 1-1over 2^v-1\=e^lambdaover 2^v-1-1\to 0$$hence the result.
answered 15 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
answered 15 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
answered 15 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
answered 15 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
16.5k3939
add a comment |
add a comment |
$begingroup$
The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).
answered 15 hours ago
user66081user66081
3,2581126
$endgroup$
add a comment |
$begingroup$
The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).
answered 15 hours ago
user66081user66081
3,2581126
$endgroup$
add a comment |
$begingroup$
The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).
answered 15 hours ago
user66081user66081
3,2581126
$endgroup$
The RHS is obviously the first two terms of the sum. For the remaining terms, replace $n!$ by $2^n$. Then, whatever $lambda$, for a sufficiently large $v$, you have a convergent geometric series (that tends to zero as $v to infty$).
answered 15 hours ago
user66081user66081
3,2581126
answered 15 hours ago
user66081user66081
3,2581126
answered 15 hours ago
user66081user66081
3,2581126
answered 15 hours ago
user66081user66081
3,2581126
3,2581126
add a comment |
add a comment |
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