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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 12 hours ago
Jwan622Jwan622
2,30011632
$endgroup$
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 12 hours ago
Jwan622Jwan622
2,30011632
$endgroup$
$begingroup$
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
12 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 12 hours ago
Jwan622Jwan622
2,30011632
$endgroup$
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 12 hours ago
Jwan622Jwan622
2,30011632
asked 12 hours ago
Jwan622Jwan622
2,30011632
edited 12 hours ago
Jwan622
asked 12 hours ago
Jwan622Jwan622
2,30011632
asked 12 hours ago
Jwan622Jwan622
2,30011632
asked 12 hours ago
Jwan622Jwan622
2,30011632
2,30011632
$begingroup$
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
12 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
12 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
12 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
12 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 12 hours ago
StackTDStackTD
24.1k2254
$endgroup$
add a comment |
$begingroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 12 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 12 hours ago
Matthew MasarikMatthew Masarik
211
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
11 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
11 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
11 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 12 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 12 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 12 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 12 hours ago
StackTDStackTD
24.1k2254
answered 12 hours ago
StackTDStackTD
24.1k2254
answered 12 hours ago
StackTDStackTD
24.1k2254
answered 12 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 12 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 12 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 12 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
$endgroup$
Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.
answered 12 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
edited 11 hours ago
answered 12 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered 12 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.1k42867
answered 12 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 12 hours ago
Matthew MasarikMatthew Masarik
211
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
11 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
11 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
11 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 12 hours ago
Matthew MasarikMatthew Masarik
211
$endgroup$
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
11 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
11 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
11 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 12 hours ago
Matthew MasarikMatthew Masarik
211
$endgroup$
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 12 hours ago
Matthew MasarikMatthew Masarik
211
answered 12 hours ago
Matthew MasarikMatthew Masarik
211
answered 12 hours ago
Matthew MasarikMatthew Masarik
211
answered 12 hours ago
Matthew MasarikMatthew Masarik
211
211
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
11 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
11 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
11 hours ago
add a comment |
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
11 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
11 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
11 hours ago
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
11 hours ago
$begingroup$
How did you get to here?
$endgroup$
– Jwan622
11 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
11 hours ago
$begingroup$
Multiply by $fracn^1.5n^1.5$.
$endgroup$
– Matthew Masarik
11 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
11 hours ago
$begingroup$
Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
$endgroup$
– Jwan622
11 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
12 hours ago