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Does this power sequence converge or diverge? If it converges, what is the limit?
The Next CEO of Stack OverflowWhy does this pattern fail (sometimes) for the continued fraction convergents of $sqrt2$?Does this sequence always give a square number?Does the sequence $A_n = fracsin (n)sqrtn$ converge to $0$?Do this series converge or diverge?Does this sequence converge $a_n=frac 3^n+25^n $Given: $sum n a_n$ is convergent. To prove: The sequence $a_n$ convergesHow do you find the value of $sum_r=1^infty frac6^r(3^r-2^r)(3^r+1 - 2^r+1) $?Confirming that the sequence $a_n = fracsqrtncosnsqrtn^3-1$ convergesDoes this non-monotonic sequence converge?Finding the limit of a sequence involving n-th roots.
$begingroup$
Say I have this sequence:
$$a_n = fracn^2sqrtn^3 + 4n$$
Again, I don't think I can divide the numerator and denominator by $n^1.5$... that seems like it complicates things. What else can I do?
I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?
Is this valid:
$$a_n = frac1sqrtfracn^3n^4 + frac4n$$
sequences-and-series
asked 2 days ago
Jwan622Jwan622
2,32211632
$endgroup$
add a comment |
$begingroup$
Say I have this sequence:
$$a_n = fracn^2sqrtn^3 + 4n$$
Again, I don't think I can divide the numerator and denominator by $n^1.5$... that seems like it complicates things. What else can I do?
I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?
Is this valid:
$$a_n = frac1sqrtfracn^3n^4 + frac4n$$
sequences-and-series
asked 2 days ago
Jwan622Jwan622
2,32211632
$endgroup$
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
2 days ago
add a comment |
$begingroup$
Say I have this sequence:
$$a_n = fracn^2sqrtn^3 + 4n$$
Again, I don't think I can divide the numerator and denominator by $n^1.5$... that seems like it complicates things. What else can I do?
I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?
Is this valid:
$$a_n = frac1sqrtfracn^3n^4 + frac4n$$
sequences-and-series
asked 2 days ago
Jwan622Jwan622
2,32211632
$endgroup$
Say I have this sequence:
$$a_n = fracn^2sqrtn^3 + 4n$$
Again, I don't think I can divide the numerator and denominator by $n^1.5$... that seems like it complicates things. What else can I do?
I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?
Is this valid:
$$a_n = frac1sqrtfracn^3n^4 + frac4n$$
sequences-and-series
sequences-and-series
asked 2 days ago
Jwan622Jwan622
2,32211632
asked 2 days ago
Jwan622Jwan622
2,32211632
edited 2 days ago
Jwan622
asked 2 days ago
Jwan622Jwan622
2,32211632
asked 2 days ago
Jwan622Jwan622
2,32211632
asked 2 days ago
Jwan622Jwan622
2,32211632
2,32211632
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
2 days ago
add a comment |
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
2 days ago
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
2 days ago
$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 2 days ago
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
add a comment |
$begingroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 2 days ago
MatthewPeterMatthewPeter
191
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
2 days ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– MatthewPeter
2 days ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
2 days ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 2 days ago
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 2 days ago
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 2 days ago
StackTDStackTD
24.3k2254
$endgroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 2 days ago
StackTDStackTD
24.3k2254
answered 2 days ago
StackTDStackTD
24.3k2254
answered 2 days ago
StackTDStackTD
24.3k2254
answered 2 days ago
StackTDStackTD
24.3k2254
24.3k2254
add a comment |
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
edited 2 days ago
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
78.2k42867
add a comment |
add a comment |
$begingroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 2 days ago
MatthewPeterMatthewPeter
191
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
2 days ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– MatthewPeter
2 days ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
2 days ago
add a comment |
$begingroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 2 days ago
MatthewPeterMatthewPeter
191
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
2 days ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– MatthewPeter
2 days ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
2 days ago
add a comment |
$begingroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 2 days ago
MatthewPeterMatthewPeter
191
$endgroup$
We have:
$$a_n = fracsqrtn sqrt1 + frac4n^2$$
You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.
answered 2 days ago
MatthewPeterMatthewPeter
191
answered 2 days ago
MatthewPeterMatthewPeter
191
answered 2 days ago
MatthewPeterMatthewPeter
191
answered 2 days ago
MatthewPeterMatthewPeter
191
191
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
2 days ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– MatthewPeter
2 days ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
2 days ago
add a comment |
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
2 days ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– MatthewPeter
2 days ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
2 days ago
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
2 days ago
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
2 days ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– MatthewPeter
2 days ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– MatthewPeter
2 days ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
2 days ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
2 days ago
add a comment |
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$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
2 days ago