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Does this power sequence converge or diverge? If it converges, what is the limit?
Why 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 5 hours ago
Jwan622Jwan622
2,30211632
$endgroup$
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 5 hours ago
Jwan622Jwan622
2,30211632
$endgroup$
$begingroup$
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
5 hours 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 5 hours ago
Jwan622Jwan622
2,30211632
$endgroup$
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 5 hours ago
Jwan622Jwan622
2,30211632
asked 5 hours ago
Jwan622Jwan622
2,30211632
edited 5 hours ago
Jwan622
asked 5 hours ago
Jwan622Jwan622
2,30211632
asked 5 hours ago
Jwan622Jwan622
2,30211632
asked 5 hours ago
Jwan622Jwan622
2,30211632
2,30211632
$begingroup$
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
5 hours 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
5 hours 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
5 hours 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
5 hours 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 5 hours ago
StackTDStackTD
24.1k2254
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
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 5 hours ago
Matthew MasarikMatthew Masarik
111
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
4 hours 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 5 hours ago
StackTDStackTD
24.1k2254
$endgroup$
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 5 hours ago
StackTDStackTD
24.1k2254
$endgroup$
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 5 hours ago
StackTDStackTD
24.1k2254
$endgroup$
You can easily find a divergent minorant:
$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$
answered 5 hours ago
StackTDStackTD
24.1k2254
answered 5 hours ago
StackTDStackTD
24.1k2254
answered 5 hours ago
StackTDStackTD
24.1k2254
answered 5 hours ago
StackTDStackTD
24.1k2254
24.1k2254
add a comment |
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
edited 5 hours ago
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered 5 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
78.1k42867
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 5 hours ago
Matthew MasarikMatthew Masarik
111
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
4 hours 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 5 hours ago
Matthew MasarikMatthew Masarik
111
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
4 hours 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 5 hours ago
Matthew MasarikMatthew Masarik
111
$endgroup$
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 5 hours ago
Matthew MasarikMatthew Masarik
111
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
answered 5 hours ago
Matthew MasarikMatthew Masarik
111
111
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
4 hours ago
add a comment |
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
4 hours ago
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
5 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
4 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
4 hours 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
5 hours ago