Covergence/ divergence of $left[frac {1+(n+1)^c}{n^c}right]^n $ [on hold]
Prove $$left(frac {1+(n+1)^c}{n^c}right)^n longrightarrow begin{cases} infty &text{if $0<c<1$}, \ e^2 &text{if $c=1$}, \e^c &text{if $c>1$}.end{cases}$$
calculus sequences-and-series
edited Dec 26 at 1:48
Chinnapparaj R
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asked Dec 26 at 1:41
Hongyan
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put on hold as off-topic by Dave, Siong Thye Goh, Chinnapparaj R, Key Flex, kingW3 Dec 26 at 3:05
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Prove $$left(frac {1+(n+1)^c}{n^c}right)^n longrightarrow begin{cases} infty &text{if $0<c<1$}, \ e^2 &text{if $c=1$}, \e^c &text{if $c>1$}.end{cases}$$
calculus sequences-and-series
edited Dec 26 at 1:48
Chinnapparaj R
5,2751826
asked Dec 26 at 1:41
Hongyan
1297
put on hold as off-topic by Dave, Siong Thye Goh, Chinnapparaj R, Key Flex, kingW3 Dec 26 at 3:05
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dave, Siong Thye Goh, Chinnapparaj R, Key Flex, kingW3
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Prove $$left(frac {1+(n+1)^c}{n^c}right)^n longrightarrow begin{cases} infty &text{if $0<c<1$}, \ e^2 &text{if $c=1$}, \e^c &text{if $c>1$}.end{cases}$$
calculus sequences-and-series
edited Dec 26 at 1:48
Chinnapparaj R
5,2751826
asked Dec 26 at 1:41
Hongyan
1297
Prove $$left(frac {1+(n+1)^c}{n^c}right)^n longrightarrow begin{cases} infty &text{if $0<c<1$}, \ e^2 &text{if $c=1$}, \e^c &text{if $c>1$}.end{cases}$$
calculus sequences-and-series
calculus sequences-and-series
edited Dec 26 at 1:48
Chinnapparaj R
5,2751826
asked Dec 26 at 1:41
Hongyan
1297
edited Dec 26 at 1:48
Chinnapparaj R
5,2751826
asked Dec 26 at 1:41
Hongyan
1297
edited Dec 26 at 1:48
Chinnapparaj R
5,2751826
edited Dec 26 at 1:48
Chinnapparaj R
5,2751826
edited Dec 26 at 1:48
Chinnapparaj R
5,2751826
5,2751826
asked Dec 26 at 1:41
Hongyan
1297
asked Dec 26 at 1:41
Hongyan
1297
asked Dec 26 at 1:41
Hongyan
1297
1297
put on hold as off-topic by Dave, Siong Thye Goh, Chinnapparaj R, Key Flex, kingW3 Dec 26 at 3:05
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dave, Siong Thye Goh, Chinnapparaj R, Key Flex, kingW3
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 Dave, Siong Thye Goh, Chinnapparaj R, Key Flex, kingW3 Dec 26 at 3:05
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dave, Siong Thye Goh, Chinnapparaj R, Key Flex, kingW3
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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Consider the following expression:
$$
logleft[left(frac{1+(n+1)^c}{n^c}right)^nright]=nlogleft(frac{1+(n+1)^c}{n^c}right)=frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n}
$$
As $nrightarrowinfty$ the RHS tends to $0/0$, and thus we may apply L'Hopital's rule. Thus, if the limit of the following RHS exists, we have that:
$$
lim_{nrightarrowinfty}frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n} = lim_{nrightarrowinfty}frac{frac{-c((n+1)^c+n+1)}{n(n+1)((n+1)^c+1)}}{-1/n^2}
$$
You should then be able to finish from here. The only challenging part is having the stomach to take that L'Hopital's derivative.
answered Dec 26 at 1:57
ImNotTheGuy
46116
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider the following expression:
$$
logleft[left(frac{1+(n+1)^c}{n^c}right)^nright]=nlogleft(frac{1+(n+1)^c}{n^c}right)=frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n}
$$
As $nrightarrowinfty$ the RHS tends to $0/0$, and thus we may apply L'Hopital's rule. Thus, if the limit of the following RHS exists, we have that:
$$
lim_{nrightarrowinfty}frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n} = lim_{nrightarrowinfty}frac{frac{-c((n+1)^c+n+1)}{n(n+1)((n+1)^c+1)}}{-1/n^2}
$$
You should then be able to finish from here. The only challenging part is having the stomach to take that L'Hopital's derivative.
answered Dec 26 at 1:57
ImNotTheGuy
46116
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
Consider the following expression:
$$
logleft[left(frac{1+(n+1)^c}{n^c}right)^nright]=nlogleft(frac{1+(n+1)^c}{n^c}right)=frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n}
$$
As $nrightarrowinfty$ the RHS tends to $0/0$, and thus we may apply L'Hopital's rule. Thus, if the limit of the following RHS exists, we have that:
$$
lim_{nrightarrowinfty}frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n} = lim_{nrightarrowinfty}frac{frac{-c((n+1)^c+n+1)}{n(n+1)((n+1)^c+1)}}{-1/n^2}
$$
You should then be able to finish from here. The only challenging part is having the stomach to take that L'Hopital's derivative.
answered Dec 26 at 1:57
ImNotTheGuy
46116
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Consider the following expression:
$$
logleft[left(frac{1+(n+1)^c}{n^c}right)^nright]=nlogleft(frac{1+(n+1)^c}{n^c}right)=frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n}
$$
As $nrightarrowinfty$ the RHS tends to $0/0$, and thus we may apply L'Hopital's rule. Thus, if the limit of the following RHS exists, we have that:
$$
lim_{nrightarrowinfty}frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n} = lim_{nrightarrowinfty}frac{frac{-c((n+1)^c+n+1)}{n(n+1)((n+1)^c+1)}}{-1/n^2}
$$
You should then be able to finish from here. The only challenging part is having the stomach to take that L'Hopital's derivative.
answered Dec 26 at 1:57
ImNotTheGuy
46116
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the following expression:
$$
logleft[left(frac{1+(n+1)^c}{n^c}right)^nright]=nlogleft(frac{1+(n+1)^c}{n^c}right)=frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n}
$$
As $nrightarrowinfty$ the RHS tends to $0/0$, and thus we may apply L'Hopital's rule. Thus, if the limit of the following RHS exists, we have that:
$$
lim_{nrightarrowinfty}frac{logleft(frac{1+(n+1)^c}{n^c}right)}{1/n} = lim_{nrightarrowinfty}frac{frac{-c((n+1)^c+n+1)}{n(n+1)((n+1)^c+1)}}{-1/n^2}
$$
You should then be able to finish from here. The only challenging part is having the stomach to take that L'Hopital's derivative.
answered Dec 26 at 1:57
ImNotTheGuy
46116
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 26 at 1:57
ImNotTheGuy
46116
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 26 at 1:57
ImNotTheGuy
46116
answered Dec 26 at 1:57
ImNotTheGuy
46116
46116
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
