Double sum misunderstanding - $sum_n=2^infty frac1n! sum_k=0^infty frac(k+2)cdots(k+n)k! z^n$. [on hold]

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up vote
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I stuck on finding first two coefficients for the double sum:

$$sum_n=2^infty dfrac1n! sum_k=0^infty dfrac(k+2)cdots(k+n)k! z^n$$
first two - $n=2$ and $3$.

e.g. for n=2 the answer is $frac32$, I don't know how can I get it.

calculus

share|cite|improve this question

edited Aug 1 at 17:31

user 108128

18.8k41544

asked Aug 1 at 15:53

Metso

886

put on hold as off-topic by user21820, amWhy, John Ma, Xander Henderson, Jendrik Stelzner 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, John Ma, Xander Henderson, Jendrik Stelzner

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you asked to write these coefficients without sums involved. Otherwise it should only be $n=2~:~fracz^22sum_k=0^inftyfrack+2k!$ and $n=3~:~fracz^36sum_k=0^inftyfrac(k+2)(k+3)k!$. Maybe one can somehow use the know series $e=sum_k=0^inftyfrac1k!$.
  – mrtaurho
  Aug 1 at 16:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i want to calculate coefficients for $z^2$ and $z^3$. I guess the first is $frac32$, but I don't know how to proceed
  – Metso
  Aug 1 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  When $n=2$ the inner sum is $$sum_k=0^inftyk+2over k!$$ Can you explain where you have difficulty evaluating this? Please put your response in the body of the question, not in a comment?
  – saulspatz
  Aug 1 at 16:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The coefficient of $z^2$ is $3eover2$ not $3over2$
  – saulspatz
  Aug 1 at 16:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Think about Taylor series, e.g. en.wikipedia.org/wiki/Taylor_series .
  – user90369
  Aug 1 at 16:27
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

1

I stuck on finding first two coefficients for the double sum:

$$sum_n=2^infty dfrac1n! sum_k=0^infty dfrac(k+2)cdots(k+n)k! z^n$$
first two - $n=2$ and $3$.

e.g. for n=2 the answer is $frac32$, I don't know how can I get it.

calculus

share|cite|improve this question

edited Aug 1 at 17:31

user 108128

18.8k41544

asked Aug 1 at 15:53

Metso

886

put on hold as off-topic by user21820, amWhy, John Ma, Xander Henderson, Jendrik Stelzner 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, John Ma, Xander Henderson, Jendrik Stelzner

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you asked to write these coefficients without sums involved. Otherwise it should only be $n=2~:~fracz^22sum_k=0^inftyfrack+2k!$ and $n=3~:~fracz^36sum_k=0^inftyfrac(k+2)(k+3)k!$. Maybe one can somehow use the know series $e=sum_k=0^inftyfrac1k!$.
  – mrtaurho
  Aug 1 at 16:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i want to calculate coefficients for $z^2$ and $z^3$. I guess the first is $frac32$, but I don't know how to proceed
  – Metso
  Aug 1 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  When $n=2$ the inner sum is $$sum_k=0^inftyk+2over k!$$ Can you explain where you have difficulty evaluating this? Please put your response in the body of the question, not in a comment?
  – saulspatz
  Aug 1 at 16:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The coefficient of $z^2$ is $3eover2$ not $3over2$
  – saulspatz
  Aug 1 at 16:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Think about Taylor series, e.g. en.wikipedia.org/wiki/Taylor_series .
  – user90369
  Aug 1 at 16:27
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

1

up vote
-2
down vote

favorite

1

1

I stuck on finding first two coefficients for the double sum:

$$sum_n=2^infty dfrac1n! sum_k=0^infty dfrac(k+2)cdots(k+n)k! z^n$$
first two - $n=2$ and $3$.

e.g. for n=2 the answer is $frac32$, I don't know how can I get it.

calculus

share|cite|improve this question

edited Aug 1 at 17:31

user 108128

18.8k41544

asked Aug 1 at 15:53

Metso

886

I stuck on finding first two coefficients for the double sum:

$$sum_n=2^infty dfrac1n! sum_k=0^infty dfrac(k+2)cdots(k+n)k! z^n$$
first two - $n=2$ and $3$.

e.g. for n=2 the answer is $frac32$, I don't know how can I get it.

calculus

share|cite|improve this question

edited Aug 1 at 17:31

user 108128

18.8k41544

asked Aug 1 at 15:53

Metso

886

share|cite|improve this question

share|cite|improve this question

edited Aug 1 at 17:31

user 108128

18.8k41544

edited Aug 1 at 17:31

user 108128

18.8k41544

edited Aug 1 at 17:31

user 108128

18.8k41544

18.8k41544

asked Aug 1 at 15:53

Metso

886

asked Aug 1 at 15:53

Metso

886

asked Aug 1 at 15:53

Metso

886

886

put on hold as off-topic by user21820, amWhy, John Ma, Xander Henderson, Jendrik Stelzner 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, John Ma, Xander Henderson, Jendrik Stelzner

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by user21820, amWhy, John Ma, Xander Henderson, Jendrik Stelzner 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, John Ma, Xander Henderson, Jendrik Stelzner

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you asked to write these coefficients without sums involved. Otherwise it should only be $n=2~:~fracz^22sum_k=0^inftyfrack+2k!$ and $n=3~:~fracz^36sum_k=0^inftyfrac(k+2)(k+3)k!$. Maybe one can somehow use the know series $e=sum_k=0^inftyfrac1k!$.
  – mrtaurho
  Aug 1 at 16:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i want to calculate coefficients for $z^2$ and $z^3$. I guess the first is $frac32$, but I don't know how to proceed
  – Metso
  Aug 1 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  When $n=2$ the inner sum is $$sum_k=0^inftyk+2over k!$$ Can you explain where you have difficulty evaluating this? Please put your response in the body of the question, not in a comment?
  – saulspatz
  Aug 1 at 16:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The coefficient of $z^2$ is $3eover2$ not $3over2$
  – saulspatz
  Aug 1 at 16:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Think about Taylor series, e.g. en.wikipedia.org/wiki/Taylor_series .
  – user90369
  Aug 1 at 16:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Are you asked to write these coefficients without sums involved. Otherwise it should only be $n=2~:~fracz^22sum_k=0^inftyfrack+2k!$ and $n=3~:~fracz^36sum_k=0^inftyfrac(k+2)(k+3)k!$. Maybe one can somehow use the know series $e=sum_k=0^inftyfrac1k!$.
  – mrtaurho
  Aug 1 at 16:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i want to calculate coefficients for $z^2$ and $z^3$. I guess the first is $frac32$, but I don't know how to proceed
  – Metso
  Aug 1 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  When $n=2$ the inner sum is $$sum_k=0^inftyk+2over k!$$ Can you explain where you have difficulty evaluating this? Please put your response in the body of the question, not in a comment?
  – saulspatz
  Aug 1 at 16:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The coefficient of $z^2$ is $3eover2$ not $3over2$
  – saulspatz
  Aug 1 at 16:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Think about Taylor series, e.g. en.wikipedia.org/wiki/Taylor_series .
  – user90369
  Aug 1 at 16:27
  

  
  
  
  

Are you asked to write these coefficients without sums involved. Otherwise it should only be $n=2~:~fracz^22sum_k=0^inftyfrack+2k!$ and $n=3~:~fracz^36sum_k=0^inftyfrac(k+2)(k+3)k!$. Maybe one can somehow use the know series $e=sum_k=0^inftyfrac1k!$.
– mrtaurho
Aug 1 at 16:07

Are you asked to write these coefficients without sums involved. Otherwise it should only be $n=2~:~fracz^22sum_k=0^inftyfrack+2k!$ and $n=3~:~fracz^36sum_k=0^inftyfrac(k+2)(k+3)k!$. Maybe one can somehow use the know series $e=sum_k=0^inftyfrac1k!$.
– mrtaurho
Aug 1 at 16:07

i want to calculate coefficients for $z^2$ and $z^3$. I guess the first is $frac32$, but I don't know how to proceed
– Metso
Aug 1 at 16:12

i want to calculate coefficients for $z^2$ and $z^3$. I guess the first is $frac32$, but I don't know how to proceed
– Metso
Aug 1 at 16:12

1

1

When $n=2$ the inner sum is $$sum_k=0^inftyk+2over k!$$ Can you explain where you have difficulty evaluating this? Please put your response in the body of the question, not in a comment?
– saulspatz
Aug 1 at 16:13

When $n=2$ the inner sum is $$sum_k=0^inftyk+2over k!$$ Can you explain where you have difficulty evaluating this? Please put your response in the body of the question, not in a comment?
– saulspatz
Aug 1 at 16:13

1

1

The coefficient of $z^2$ is $3eover2$ not $3over2$
– saulspatz
Aug 1 at 16:18

The coefficient of $z^2$ is $3eover2$ not $3over2$
– saulspatz
Aug 1 at 16:18

Think about Taylor series, e.g. en.wikipedia.org/wiki/Taylor_series .
– user90369
Aug 1 at 16:27

Think about Taylor series, e.g. en.wikipedia.org/wiki/Taylor_series .
– user90369
Aug 1 at 16:27

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

For the $n=2$ case, the coefficient of $z^2$ is $$
frac12!sum_k=0^inftyk+2over k!=frac12sum_k=1^infty1over(k-1)!+sum_k=0^infty1over k!=
frac12sum_k=0^infty1over k!+sum_k=0^infty1over k!=3eover 2
$$

Now try to do the same sort of thing for the $n=3$ case. Hint: Rewrite the numerator $(k+2)(k+3)$ as $k(k+1)+4k+6.$

share|cite|improve this answer

answered Aug 1 at 16:26

saulspatz

10.5k21324

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you very much
  – Metso
  Aug 1 at 16:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have thought about this approach by myself but I am not sure about the index shift after splitting up the sum. Clearly the term $frac1(k-1)!$ makes no sense for $k=0$ but how can you justify that you lift up the index by $1$?
  – mrtaurho
  Aug 1 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When $k=0$ the first term is $0$ so we're really summing from $k=1$. Then you cancel the $k$. I combined two steps: change the lower index to $k=1;$ cancel $k$ from numerator and denominator.
  – saulspatz
  Aug 1 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not asked the question so I was not able to do so ^^'. Nevertheless I am still concerned about this step.
  – mrtaurho
  Aug 1 at 16:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mrtaurho Oops, sorry. I didn't look at the signature, I guess.
  – saulspatz
  Aug 1 at 16:38
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

For the $n=2$ case, the coefficient of $z^2$ is $$
frac12!sum_k=0^inftyk+2over k!=frac12sum_k=1^infty1over(k-1)!+sum_k=0^infty1over k!=
frac12sum_k=0^infty1over k!+sum_k=0^infty1over k!=3eover 2
$$

Now try to do the same sort of thing for the $n=3$ case. Hint: Rewrite the numerator $(k+2)(k+3)$ as $k(k+1)+4k+6.$

share|cite|improve this answer

answered Aug 1 at 16:26

saulspatz

10.5k21324

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you very much
  – Metso
  Aug 1 at 16:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have thought about this approach by myself but I am not sure about the index shift after splitting up the sum. Clearly the term $frac1(k-1)!$ makes no sense for $k=0$ but how can you justify that you lift up the index by $1$?
  – mrtaurho
  Aug 1 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When $k=0$ the first term is $0$ so we're really summing from $k=1$. Then you cancel the $k$. I combined two steps: change the lower index to $k=1;$ cancel $k$ from numerator and denominator.
  – saulspatz
  Aug 1 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not asked the question so I was not able to do so ^^'. Nevertheless I am still concerned about this step.
  – mrtaurho
  Aug 1 at 16:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mrtaurho Oops, sorry. I didn't look at the signature, I guess.
  – saulspatz
  Aug 1 at 16:38
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

For the $n=2$ case, the coefficient of $z^2$ is $$
frac12!sum_k=0^inftyk+2over k!=frac12sum_k=1^infty1over(k-1)!+sum_k=0^infty1over k!=
frac12sum_k=0^infty1over k!+sum_k=0^infty1over k!=3eover 2
$$

Now try to do the same sort of thing for the $n=3$ case. Hint: Rewrite the numerator $(k+2)(k+3)$ as $k(k+1)+4k+6.$

share|cite|improve this answer

answered Aug 1 at 16:26

saulspatz

10.5k21324

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you very much
  – Metso
  Aug 1 at 16:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have thought about this approach by myself but I am not sure about the index shift after splitting up the sum. Clearly the term $frac1(k-1)!$ makes no sense for $k=0$ but how can you justify that you lift up the index by $1$?
  – mrtaurho
  Aug 1 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When $k=0$ the first term is $0$ so we're really summing from $k=1$. Then you cancel the $k$. I combined two steps: change the lower index to $k=1;$ cancel $k$ from numerator and denominator.
  – saulspatz
  Aug 1 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not asked the question so I was not able to do so ^^'. Nevertheless I am still concerned about this step.
  – mrtaurho
  Aug 1 at 16:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mrtaurho Oops, sorry. I didn't look at the signature, I guess.
  – saulspatz
  Aug 1 at 16:38
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

For the $n=2$ case, the coefficient of $z^2$ is $$
frac12!sum_k=0^inftyk+2over k!=frac12sum_k=1^infty1over(k-1)!+sum_k=0^infty1over k!=
frac12sum_k=0^infty1over k!+sum_k=0^infty1over k!=3eover 2
$$

Now try to do the same sort of thing for the $n=3$ case. Hint: Rewrite the numerator $(k+2)(k+3)$ as $k(k+1)+4k+6.$

share|cite|improve this answer

answered Aug 1 at 16:26

saulspatz

10.5k21324

For the $n=2$ case, the coefficient of $z^2$ is $$
frac12!sum_k=0^inftyk+2over k!=frac12sum_k=1^infty1over(k-1)!+sum_k=0^infty1over k!=
frac12sum_k=0^infty1over k!+sum_k=0^infty1over k!=3eover 2
$$

Now try to do the same sort of thing for the $n=3$ case. Hint: Rewrite the numerator $(k+2)(k+3)$ as $k(k+1)+4k+6.$

share|cite|improve this answer

answered Aug 1 at 16:26

saulspatz

10.5k21324

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 16:26

saulspatz

10.5k21324

answered Aug 1 at 16:26

saulspatz

10.5k21324

answered Aug 1 at 16:26

saulspatz

10.5k21324

10.5k21324

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you very much
  – Metso
  Aug 1 at 16:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have thought about this approach by myself but I am not sure about the index shift after splitting up the sum. Clearly the term $frac1(k-1)!$ makes no sense for $k=0$ but how can you justify that you lift up the index by $1$?
  – mrtaurho
  Aug 1 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When $k=0$ the first term is $0$ so we're really summing from $k=1$. Then you cancel the $k$. I combined two steps: change the lower index to $k=1;$ cancel $k$ from numerator and denominator.
  – saulspatz
  Aug 1 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not asked the question so I was not able to do so ^^'. Nevertheless I am still concerned about this step.
  – mrtaurho
  Aug 1 at 16:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mrtaurho Oops, sorry. I didn't look at the signature, I guess.
  – saulspatz
  Aug 1 at 16:38
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thank you very much
  – Metso
  Aug 1 at 16:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have thought about this approach by myself but I am not sure about the index shift after splitting up the sum. Clearly the term $frac1(k-1)!$ makes no sense for $k=0$ but how can you justify that you lift up the index by $1$?
  – mrtaurho
  Aug 1 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When $k=0$ the first term is $0$ so we're really summing from $k=1$. Then you cancel the $k$. I combined two steps: change the lower index to $k=1;$ cancel $k$ from numerator and denominator.
  – saulspatz
  Aug 1 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did not asked the question so I was not able to do so ^^'. Nevertheless I am still concerned about this step.
  – mrtaurho
  Aug 1 at 16:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mrtaurho Oops, sorry. I didn't look at the signature, I guess.
  – saulspatz
  Aug 1 at 16:38
  

  
  
  
  

thank you very much
– Metso
Aug 1 at 16:27

thank you very much
– Metso
Aug 1 at 16:27

I have thought about this approach by myself but I am not sure about the index shift after splitting up the sum. Clearly the term $frac1(k-1)!$ makes no sense for $k=0$ but how can you justify that you lift up the index by $1$?
– mrtaurho
Aug 1 at 16:29

I have thought about this approach by myself but I am not sure about the index shift after splitting up the sum. Clearly the term $frac1(k-1)!$ makes no sense for $k=0$ but how can you justify that you lift up the index by $1$?
– mrtaurho
Aug 1 at 16:29

When $k=0$ the first term is $0$ so we're really summing from $k=1$. Then you cancel the $k$. I combined two steps: change the lower index to $k=1;$ cancel $k$ from numerator and denominator.
– saulspatz
Aug 1 at 16:33

When $k=0$ the first term is $0$ so we're really summing from $k=1$. Then you cancel the $k$. I combined two steps: change the lower index to $k=1;$ cancel $k$ from numerator and denominator.
– saulspatz
Aug 1 at 16:33

I did not asked the question so I was not able to do so ^^'. Nevertheless I am still concerned about this step.
– mrtaurho
Aug 1 at 16:36

I did not asked the question so I was not able to do so ^^'. Nevertheless I am still concerned about this step.
– mrtaurho
Aug 1 at 16:36

@mrtaurho Oops, sorry. I didn't look at the signature, I guess.
– saulspatz
Aug 1 at 16:38

@mrtaurho Oops, sorry. I didn't look at the signature, I guess.
– saulspatz
Aug 1 at 16:38

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