Limit of sequences - including infinite limits [closed]
I am in need of finding limits for the following sequences:
$$a_n = bigg(1+ frac1n bigg)^{n^2} $$
$$b_n = (-1)^n (sqrt[n]n)$$
$$c_n = 2^n - bigg(2+ frac1n bigg)^n $$
$$d_n = bigg(sqrt[n]n + {sqrt[n]nover n} bigg)^n $$
They can either converge to a real number, diverge to $ pminfty$, or simply diverge.
So I managed to conclude that $a_n$ diverges to $+infty$, $b_n$ diverges, but I am quite stuck with $c_n$ and $d_n$.
How should I approach these sequences?
real-analysis calculus sequences-and-series limits
edited Dec 9 at 8:36
Gaby Alfonso
676315
asked Dec 9 at 8:11
Tegernako
756
closed as off-topic by RRL, amWhy, DRF, Cesareo, user21820 Dec 10 at 19:15
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." – RRL, amWhy, DRF, Cesareo, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I am in need of finding limits for the following sequences:
$$a_n = bigg(1+ frac1n bigg)^{n^2} $$
$$b_n = (-1)^n (sqrt[n]n)$$
$$c_n = 2^n - bigg(2+ frac1n bigg)^n $$
$$d_n = bigg(sqrt[n]n + {sqrt[n]nover n} bigg)^n $$
They can either converge to a real number, diverge to $ pminfty$, or simply diverge.
So I managed to conclude that $a_n$ diverges to $+infty$, $b_n$ diverges, but I am quite stuck with $c_n$ and $d_n$.
How should I approach these sequences?
real-analysis calculus sequences-and-series limits
edited Dec 9 at 8:36
Gaby Alfonso
676315
asked Dec 9 at 8:11
Tegernako
756
closed as off-topic by RRL, amWhy, DRF, Cesareo, user21820 Dec 10 at 19:15
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." – RRL, amWhy, DRF, Cesareo, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.
2
For $c_n$, consider trying to pull a factor of $2^n$ out of the right hand side, then see if you can figure out a limit for the term in parentheses.
– Steven Stadnicki
Dec 9 at 8:17
add a comment |
I am in need of finding limits for the following sequences:
$$a_n = bigg(1+ frac1n bigg)^{n^2} $$
$$b_n = (-1)^n (sqrt[n]n)$$
$$c_n = 2^n - bigg(2+ frac1n bigg)^n $$
$$d_n = bigg(sqrt[n]n + {sqrt[n]nover n} bigg)^n $$
They can either converge to a real number, diverge to $ pminfty$, or simply diverge.
So I managed to conclude that $a_n$ diverges to $+infty$, $b_n$ diverges, but I am quite stuck with $c_n$ and $d_n$.
How should I approach these sequences?
real-analysis calculus sequences-and-series limits
edited Dec 9 at 8:36
Gaby Alfonso
676315
asked Dec 9 at 8:11
Tegernako
756
I am in need of finding limits for the following sequences:
$$a_n = bigg(1+ frac1n bigg)^{n^2} $$
$$b_n = (-1)^n (sqrt[n]n)$$
$$c_n = 2^n - bigg(2+ frac1n bigg)^n $$
$$d_n = bigg(sqrt[n]n + {sqrt[n]nover n} bigg)^n $$
They can either converge to a real number, diverge to $ pminfty$, or simply diverge.
So I managed to conclude that $a_n$ diverges to $+infty$, $b_n$ diverges, but I am quite stuck with $c_n$ and $d_n$.
How should I approach these sequences?
real-analysis calculus sequences-and-series limits
real-analysis calculus sequences-and-series limits
edited Dec 9 at 8:36
Gaby Alfonso
676315
asked Dec 9 at 8:11
Tegernako
756
edited Dec 9 at 8:36
Gaby Alfonso
676315
asked Dec 9 at 8:11
Tegernako
756
edited Dec 9 at 8:36
Gaby Alfonso
676315
edited Dec 9 at 8:36
Gaby Alfonso
676315
edited Dec 9 at 8:36
Gaby Alfonso
676315
676315
asked Dec 9 at 8:11
Tegernako
756
asked Dec 9 at 8:11
Tegernako
756
asked Dec 9 at 8:11
Tegernako
756
756
closed as off-topic by RRL, amWhy, DRF, Cesareo, user21820 Dec 10 at 19:15
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." – RRL, amWhy, DRF, Cesareo, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, amWhy, DRF, Cesareo, user21820 Dec 10 at 19:15
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." – RRL, amWhy, DRF, Cesareo, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.
2
For $c_n$, consider trying to pull a factor of $2^n$ out of the right hand side, then see if you can figure out a limit for the term in parentheses.
– Steven Stadnicki
Dec 9 at 8:17
add a comment |
2
For $c_n$, consider trying to pull a factor of $2^n$ out of the right hand side, then see if you can figure out a limit for the term in parentheses.
– Steven Stadnicki
Dec 9 at 8:17
2
2
For $c_n$, consider trying to pull a factor of $2^n$ out of the right hand side, then see if you can figure out a limit for the term in parentheses.
– Steven Stadnicki
Dec 9 at 8:17
For $c_n$, consider trying to pull a factor of $2^n$ out of the right hand side, then see if you can figure out a limit for the term in parentheses.
– Steven Stadnicki
Dec 9 at 8:17
add a comment |
1 Answer
1
active
oldest
votes
HINT
For the first two you are right
$$a_n = bigg(1+ frac1n bigg)^{n^2}=left[ bigg(1+ frac1n bigg)^{n}right]^n toinfty$$
$$b_n = (-1)^n (sqrt[n]n) to begin{cases}1quad n=2k\-1quad n=2k+1 end{cases}$$
For the third one use that
$$c_n = 2^n - bigg(2+ frac1n bigg)^n= 2^n - 2^nbigg(1+ frac1{2n} bigg)^n$$
and for the last one
$$d_n = bigg(sqrt[n]n + {sqrt[n]nover n} bigg)^n= (sqrt[n]n)^nbigg(1 + {1over n} bigg)^n = nbigg(1 + {1over n} bigg)^n$$
answered Dec 9 at 8:32
gimusi
1
So for $c_n$, both expressions will go to infinity as n gets larger, and I will have $infty - infty$, which is not defined, thus $c_n$ diverge?
– Tegernako
Dec 9 at 8:50
1
@Tegernako For $c_n$, the indeterminate form $infty-infty$ doesn't allow to conclude, we need a step more. What about factorin out $2^n$?
– gimusi
Dec 9 at 8:54
Once I factor out $2^n$ , As $n$ increases, the expressions inside will go to $(1-e)$ ? And therefore we will diverge into $-infty$ ?
– Tegernako
Dec 9 at 8:58
1
@Tegernako Maybe $(1-sqrt e)$? But the idea is that exactly!
– gimusi
Dec 9 at 9:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
For the first two you are right
$$a_n = bigg(1+ frac1n bigg)^{n^2}=left[ bigg(1+ frac1n bigg)^{n}right]^n toinfty$$
$$b_n = (-1)^n (sqrt[n]n) to begin{cases}1quad n=2k\-1quad n=2k+1 end{cases}$$
For the third one use that
$$c_n = 2^n - bigg(2+ frac1n bigg)^n= 2^n - 2^nbigg(1+ frac1{2n} bigg)^n$$
and for the last one
$$d_n = bigg(sqrt[n]n + {sqrt[n]nover n} bigg)^n= (sqrt[n]n)^nbigg(1 + {1over n} bigg)^n = nbigg(1 + {1over n} bigg)^n$$
answered Dec 9 at 8:32
gimusi
1
So for $c_n$, both expressions will go to infinity as n gets larger, and I will have $infty - infty$, which is not defined, thus $c_n$ diverge?
– Tegernako
Dec 9 at 8:50
1
@Tegernako For $c_n$, the indeterminate form $infty-infty$ doesn't allow to conclude, we need a step more. What about factorin out $2^n$?
– gimusi
Dec 9 at 8:54
Once I factor out $2^n$ , As $n$ increases, the expressions inside will go to $(1-e)$ ? And therefore we will diverge into $-infty$ ?
– Tegernako
Dec 9 at 8:58
1
@Tegernako Maybe $(1-sqrt e)$? But the idea is that exactly!
– gimusi
Dec 9 at 9:00
add a comment |
HINT
For the first two you are right
$$a_n = bigg(1+ frac1n bigg)^{n^2}=left[ bigg(1+ frac1n bigg)^{n}right]^n toinfty$$
$$b_n = (-1)^n (sqrt[n]n) to begin{cases}1quad n=2k\-1quad n=2k+1 end{cases}$$
For the third one use that
$$c_n = 2^n - bigg(2+ frac1n bigg)^n= 2^n - 2^nbigg(1+ frac1{2n} bigg)^n$$
and for the last one
$$d_n = bigg(sqrt[n]n + {sqrt[n]nover n} bigg)^n= (sqrt[n]n)^nbigg(1 + {1over n} bigg)^n = nbigg(1 + {1over n} bigg)^n$$
answered Dec 9 at 8:32
gimusi
1
So for $c_n$, both expressions will go to infinity as n gets larger, and I will have $infty - infty$, which is not defined, thus $c_n$ diverge?
– Tegernako
Dec 9 at 8:50
1
@Tegernako For $c_n$, the indeterminate form $infty-infty$ doesn't allow to conclude, we need a step more. What about factorin out $2^n$?
– gimusi
Dec 9 at 8:54
Once I factor out $2^n$ , As $n$ increases, the expressions inside will go to $(1-e)$ ? And therefore we will diverge into $-infty$ ?
– Tegernako
Dec 9 at 8:58
1
@Tegernako Maybe $(1-sqrt e)$? But the idea is that exactly!
– gimusi
Dec 9 at 9:00
add a comment |
HINT
For the first two you are right
$$a_n = bigg(1+ frac1n bigg)^{n^2}=left[ bigg(1+ frac1n bigg)^{n}right]^n toinfty$$
$$b_n = (-1)^n (sqrt[n]n) to begin{cases}1quad n=2k\-1quad n=2k+1 end{cases}$$
For the third one use that
$$c_n = 2^n - bigg(2+ frac1n bigg)^n= 2^n - 2^nbigg(1+ frac1{2n} bigg)^n$$
and for the last one
$$d_n = bigg(sqrt[n]n + {sqrt[n]nover n} bigg)^n= (sqrt[n]n)^nbigg(1 + {1over n} bigg)^n = nbigg(1 + {1over n} bigg)^n$$
answered Dec 9 at 8:32
gimusi
1
HINT
For the first two you are right
$$a_n = bigg(1+ frac1n bigg)^{n^2}=left[ bigg(1+ frac1n bigg)^{n}right]^n toinfty$$
$$b_n = (-1)^n (sqrt[n]n) to begin{cases}1quad n=2k\-1quad n=2k+1 end{cases}$$
For the third one use that
$$c_n = 2^n - bigg(2+ frac1n bigg)^n= 2^n - 2^nbigg(1+ frac1{2n} bigg)^n$$
and for the last one
$$d_n = bigg(sqrt[n]n + {sqrt[n]nover n} bigg)^n= (sqrt[n]n)^nbigg(1 + {1over n} bigg)^n = nbigg(1 + {1over n} bigg)^n$$
answered Dec 9 at 8:32
gimusi
1
answered Dec 9 at 8:32
gimusi
1
answered Dec 9 at 8:32
gimusi
1
answered Dec 9 at 8:32
gimusi
1
1
So for $c_n$, both expressions will go to infinity as n gets larger, and I will have $infty - infty$, which is not defined, thus $c_n$ diverge?
– Tegernako
Dec 9 at 8:50
1
@Tegernako For $c_n$, the indeterminate form $infty-infty$ doesn't allow to conclude, we need a step more. What about factorin out $2^n$?
– gimusi
Dec 9 at 8:54
Once I factor out $2^n$ , As $n$ increases, the expressions inside will go to $(1-e)$ ? And therefore we will diverge into $-infty$ ?
– Tegernako
Dec 9 at 8:58
1
@Tegernako Maybe $(1-sqrt e)$? But the idea is that exactly!
– gimusi
Dec 9 at 9:00
add a comment |
So for $c_n$, both expressions will go to infinity as n gets larger, and I will have $infty - infty$, which is not defined, thus $c_n$ diverge?
– Tegernako
Dec 9 at 8:50
1
@Tegernako For $c_n$, the indeterminate form $infty-infty$ doesn't allow to conclude, we need a step more. What about factorin out $2^n$?
– gimusi
Dec 9 at 8:54
Once I factor out $2^n$ , As $n$ increases, the expressions inside will go to $(1-e)$ ? And therefore we will diverge into $-infty$ ?
– Tegernako
Dec 9 at 8:58
1
@Tegernako Maybe $(1-sqrt e)$? But the idea is that exactly!
– gimusi
Dec 9 at 9:00
So for $c_n$, both expressions will go to infinity as n gets larger, and I will have $infty - infty$, which is not defined, thus $c_n$ diverge?
– Tegernako
Dec 9 at 8:50
So for $c_n$, both expressions will go to infinity as n gets larger, and I will have $infty - infty$, which is not defined, thus $c_n$ diverge?
– Tegernako
Dec 9 at 8:50
1
1
@Tegernako For $c_n$, the indeterminate form $infty-infty$ doesn't allow to conclude, we need a step more. What about factorin out $2^n$?
– gimusi
Dec 9 at 8:54
@Tegernako For $c_n$, the indeterminate form $infty-infty$ doesn't allow to conclude, we need a step more. What about factorin out $2^n$?
– gimusi
Dec 9 at 8:54
Once I factor out $2^n$ , As $n$ increases, the expressions inside will go to $(1-e)$ ? And therefore we will diverge into $-infty$ ?
– Tegernako
Dec 9 at 8:58
Once I factor out $2^n$ , As $n$ increases, the expressions inside will go to $(1-e)$ ? And therefore we will diverge into $-infty$ ?
– Tegernako
Dec 9 at 8:58
1
1
@Tegernako Maybe $(1-sqrt e)$? But the idea is that exactly!
– gimusi
Dec 9 at 9:00
@Tegernako Maybe $(1-sqrt e)$? But the idea is that exactly!
– gimusi
Dec 9 at 9:00
add a comment |
2
For $c_n$, consider trying to pull a factor of $2^n$ out of the right hand side, then see if you can figure out a limit for the term in parentheses.
– Steven Stadnicki
Dec 9 at 8:17