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 9 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 9 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 9 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 9 hours ago
math studentmath student
2,39111018
asked 9 hours ago
math studentmath student
2,39111018
asked 9 hours ago
math studentmath student
2,39111018
asked 9 hours ago
math studentmath student
2,39111018
asked 9 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 9 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 9 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 9 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 9 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 9 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 9 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
answered 9 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
answered 9 hours ago
Mostafa AyazMostafa Ayaz
16.5k3939
answered 9 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 9 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 9 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 9 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 9 hours ago
user66081user66081
3,2581126
answered 9 hours ago
user66081user66081
3,2581126
answered 9 hours ago
user66081user66081
3,2581126
answered 9 hours ago
user66081user66081
3,2581126
3,2581126
add a comment |
add a comment |
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