Computation of a limit involving a series (related to Poisson distribution)
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Consider $lambda >0.$ I am reading a paper and the author states that
$$ displaystylelim_{v rightarrow +infty} sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v} = 1 + lambda$$
I tried to compute such limit but I am getting anywhere. Someone could help me?
Thanks in advance!
analysis statistics
$endgroup$
add a comment |
$begingroup$
Consider $lambda >0.$ I am reading a paper and the author states that
$$ displaystylelim_{v rightarrow +infty} sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v} = 1 + lambda$$
I tried to compute such limit but I am getting anywhere. Someone could help me?
Thanks in advance!
analysis statistics
$endgroup$
add a comment |
$begingroup$
Consider $lambda >0.$ I am reading a paper and the author states that
$$ displaystylelim_{v rightarrow +infty} sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v} = 1 + lambda$$
I tried to compute such limit but I am getting anywhere. Someone could help me?
Thanks in advance!
analysis statistics
$endgroup$
Consider $lambda >0.$ I am reading a paper and the author states that
$$ displaystylelim_{v rightarrow +infty} sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v} = 1 + lambda$$
I tried to compute such limit but I am getting anywhere. Someone could help me?
Thanks in advance!
analysis statistics
analysis statistics
asked 2 days ago
math studentmath student
2,39111018
2,39111018
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2 Answers
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$begingroup$
Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.
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$begingroup$
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 $v to infty$).
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2 Answers
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2 Answers
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$begingroup$
Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.
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add a comment |
$begingroup$
Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.
$endgroup$
add a comment |
$begingroup$
Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.
$endgroup$
Note that $$sum_{n=0}^{+infty} frac{lambda^{n}}{(n !)^v}=1+lambda+sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}$$and $$sum_{n=2}^{+infty} frac{lambda^{n}}{(n !)^v}{le sum_{n=2}^{+infty} frac{lambda^{n}}{n !}sum_{n=2}^{+infty} frac{1}{(n !)^{v-1}}\le e^{lambda}sum_{n=2}^{infty}{1over 2^{(n-1)(v-1)}}\=e^lambda{{1over 2^{v-1}}over 1-{1over 2^{v-1}}}\={e^lambdaover 2^{v-1}-1}\to 0}$$hence the result.
answered 2 days ago
Mostafa AyazMostafa Ayaz
17k3939
17k3939
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$begingroup$
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 $v to infty$).
$endgroup$
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 $v to infty$).
$endgroup$
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 $v to infty$).
$endgroup$
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 2 days ago
user66081user66081
3,2581126
3,2581126
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