Mixing
An unorthodox way of converting a recurrent sequence into a non-recurrent one
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The problem is:
A sequence is given with:
$$ a_1 = 2016, a_n+1 = sqrta_n^2 + 4 + frac1a_n^222$$
Find:
$$ (a) quad lim_n to inftya_n quad (b) lim_n to inftyfraca_n^2n quad (c) lim_n to inftyfracsum_k=1^na_kn sqrtn$$
The (a) part I was able to solve by proving that the sequence is increasing and not bounded by above. Therefore
$$ lim_n to inftya_n = +infty$$
Then for (b) I applied the Stolz theorem:
$$ lim_n to inftyfraca_n^2n = lim_n to inftyfraca_n+1^2 - a_n^2n+1-n = lim_n to infty(a_n^2 + 4 + frac1a_n^222 - a_n^2) = lim_n to infty(4 + frac1a_n^222) = 4$$
But the (c) part is where I get confused. I've tried applying the Stolz theorem there as well but no luck since the n's don't cancel out like they did in (b).
I did have one thing on my mind though but I'm not sure how formally correct it is.
Let's look at the first few sequence members:
$$a_1 = 2016, a_2 = sqrt2016^2 + 4 + frac12016^222, a_3 = sqrt2016^2 + 4 + frac12016^222+ 4 + frac1a_2^222 = sqrt2016^2 + 8 + frac12016^222 + frac1a_2^222$$
And now as we keep going we are going to get a new $$ frac1a_n^222$$ member inside for each n. So that still makes this sequence recursive. But can I simply disregard that part since it obviously always converges to zero, not matter what the value of n is? Or maybe I could say something like
$$ textLet frac1a_n^222 = Z(n) quad text Then lim_n to inftyZ(n) = 0 quad (forall n)$$
And now I rewrite my sequence as:
$$ a_n = sqrt(2016)^2 + 4(n-1) + sum_i=1^n-1Z(i) quad forall n geq 2$$
Now let's finally look at the (c) problem:
$$lim_n to inftyfracsum_k=1^na_kn sqrtn = lim_n to inftyfraca_1 + a_2 + ... +a_nn sqrtn = lim_n to inftyfrac2016 + sqrt2016^2 + 4 + Z(1) + ... + sqrt2016^2 + 4(n-1) + sumZ(n) n sqrtn$$
And now I apply Stolz theorem:
$$ = lim_n to inftyfracsqrt2016^2 + 4n + sumZ(n+1)(n+1)sqrtn+1 - nsqrtn = 0 $$
Since the top degree of $n$ is $frac12$ and the bottom one is $ frac32$.
So can anyone tell me whether this solutions is correct? If it isn't, how would you solve it? If it is correct but my process is wrong I'd also like to know where I went wrong.
Thanks in advance.
real-analysis limits convergence
asked 2 hours ago
Koy
1776
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up vote
4
down vote
favorite
The problem is:
A sequence is given with:
$$ a_1 = 2016, a_n+1 = sqrta_n^2 + 4 + frac1a_n^222$$
Find:
$$ (a) quad lim_n to inftya_n quad (b) lim_n to inftyfraca_n^2n quad (c) lim_n to inftyfracsum_k=1^na_kn sqrtn$$
The (a) part I was able to solve by proving that the sequence is increasing and not bounded by above. Therefore
$$ lim_n to inftya_n = +infty$$
Then for (b) I applied the Stolz theorem:
$$ lim_n to inftyfraca_n^2n = lim_n to inftyfraca_n+1^2 - a_n^2n+1-n = lim_n to infty(a_n^2 + 4 + frac1a_n^222 - a_n^2) = lim_n to infty(4 + frac1a_n^222) = 4$$
But the (c) part is where I get confused. I've tried applying the Stolz theorem there as well but no luck since the n's don't cancel out like they did in (b).
I did have one thing on my mind though but I'm not sure how formally correct it is.
Let's look at the first few sequence members:
$$a_1 = 2016, a_2 = sqrt2016^2 + 4 + frac12016^222, a_3 = sqrt2016^2 + 4 + frac12016^222+ 4 + frac1a_2^222 = sqrt2016^2 + 8 + frac12016^222 + frac1a_2^222$$
And now as we keep going we are going to get a new $$ frac1a_n^222$$ member inside for each n. So that still makes this sequence recursive. But can I simply disregard that part since it obviously always converges to zero, not matter what the value of n is? Or maybe I could say something like
$$ textLet frac1a_n^222 = Z(n) quad text Then lim_n to inftyZ(n) = 0 quad (forall n)$$
And now I rewrite my sequence as:
$$ a_n = sqrt(2016)^2 + 4(n-1) + sum_i=1^n-1Z(i) quad forall n geq 2$$
Now let's finally look at the (c) problem:
$$lim_n to inftyfracsum_k=1^na_kn sqrtn = lim_n to inftyfraca_1 + a_2 + ... +a_nn sqrtn = lim_n to inftyfrac2016 + sqrt2016^2 + 4 + Z(1) + ... + sqrt2016^2 + 4(n-1) + sumZ(n) n sqrtn$$
And now I apply Stolz theorem:
$$ = lim_n to inftyfracsqrt2016^2 + 4n + sumZ(n+1)(n+1)sqrtn+1 - nsqrtn = 0 $$
Since the top degree of $n$ is $frac12$ and the bottom one is $ frac32$.
So can anyone tell me whether this solutions is correct? If it isn't, how would you solve it? If it is correct but my process is wrong I'd also like to know where I went wrong.
Thanks in advance.
real-analysis limits convergence
asked 2 hours ago
Koy
1776
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The problem is:
A sequence is given with:
$$ a_1 = 2016, a_n+1 = sqrta_n^2 + 4 + frac1a_n^222$$
Find:
$$ (a) quad lim_n to inftya_n quad (b) lim_n to inftyfraca_n^2n quad (c) lim_n to inftyfracsum_k=1^na_kn sqrtn$$
The (a) part I was able to solve by proving that the sequence is increasing and not bounded by above. Therefore
$$ lim_n to inftya_n = +infty$$
Then for (b) I applied the Stolz theorem:
$$ lim_n to inftyfraca_n^2n = lim_n to inftyfraca_n+1^2 - a_n^2n+1-n = lim_n to infty(a_n^2 + 4 + frac1a_n^222 - a_n^2) = lim_n to infty(4 + frac1a_n^222) = 4$$
But the (c) part is where I get confused. I've tried applying the Stolz theorem there as well but no luck since the n's don't cancel out like they did in (b).
I did have one thing on my mind though but I'm not sure how formally correct it is.
Let's look at the first few sequence members:
$$a_1 = 2016, a_2 = sqrt2016^2 + 4 + frac12016^222, a_3 = sqrt2016^2 + 4 + frac12016^222+ 4 + frac1a_2^222 = sqrt2016^2 + 8 + frac12016^222 + frac1a_2^222$$
And now as we keep going we are going to get a new $$ frac1a_n^222$$ member inside for each n. So that still makes this sequence recursive. But can I simply disregard that part since it obviously always converges to zero, not matter what the value of n is? Or maybe I could say something like
$$ textLet frac1a_n^222 = Z(n) quad text Then lim_n to inftyZ(n) = 0 quad (forall n)$$
And now I rewrite my sequence as:
$$ a_n = sqrt(2016)^2 + 4(n-1) + sum_i=1^n-1Z(i) quad forall n geq 2$$
Now let's finally look at the (c) problem:
$$lim_n to inftyfracsum_k=1^na_kn sqrtn = lim_n to inftyfraca_1 + a_2 + ... +a_nn sqrtn = lim_n to inftyfrac2016 + sqrt2016^2 + 4 + Z(1) + ... + sqrt2016^2 + 4(n-1) + sumZ(n) n sqrtn$$
And now I apply Stolz theorem:
$$ = lim_n to inftyfracsqrt2016^2 + 4n + sumZ(n+1)(n+1)sqrtn+1 - nsqrtn = 0 $$
Since the top degree of $n$ is $frac12$ and the bottom one is $ frac32$.
So can anyone tell me whether this solutions is correct? If it isn't, how would you solve it? If it is correct but my process is wrong I'd also like to know where I went wrong.
Thanks in advance.
real-analysis limits convergence
asked 2 hours ago
Koy
1776
The problem is:
A sequence is given with:
$$ a_1 = 2016, a_n+1 = sqrta_n^2 + 4 + frac1a_n^222$$
Find:
$$ (a) quad lim_n to inftya_n quad (b) lim_n to inftyfraca_n^2n quad (c) lim_n to inftyfracsum_k=1^na_kn sqrtn$$
The (a) part I was able to solve by proving that the sequence is increasing and not bounded by above. Therefore
$$ lim_n to inftya_n = +infty$$
Then for (b) I applied the Stolz theorem:
$$ lim_n to inftyfraca_n^2n = lim_n to inftyfraca_n+1^2 - a_n^2n+1-n = lim_n to infty(a_n^2 + 4 + frac1a_n^222 - a_n^2) = lim_n to infty(4 + frac1a_n^222) = 4$$
But the (c) part is where I get confused. I've tried applying the Stolz theorem there as well but no luck since the n's don't cancel out like they did in (b).
I did have one thing on my mind though but I'm not sure how formally correct it is.
Let's look at the first few sequence members:
$$a_1 = 2016, a_2 = sqrt2016^2 + 4 + frac12016^222, a_3 = sqrt2016^2 + 4 + frac12016^222+ 4 + frac1a_2^222 = sqrt2016^2 + 8 + frac12016^222 + frac1a_2^222$$
And now as we keep going we are going to get a new $$ frac1a_n^222$$ member inside for each n. So that still makes this sequence recursive. But can I simply disregard that part since it obviously always converges to zero, not matter what the value of n is? Or maybe I could say something like
$$ textLet frac1a_n^222 = Z(n) quad text Then lim_n to inftyZ(n) = 0 quad (forall n)$$
And now I rewrite my sequence as:
$$ a_n = sqrt(2016)^2 + 4(n-1) + sum_i=1^n-1Z(i) quad forall n geq 2$$
Now let's finally look at the (c) problem:
$$lim_n to inftyfracsum_k=1^na_kn sqrtn = lim_n to inftyfraca_1 + a_2 + ... +a_nn sqrtn = lim_n to inftyfrac2016 + sqrt2016^2 + 4 + Z(1) + ... + sqrt2016^2 + 4(n-1) + sumZ(n) n sqrtn$$
And now I apply Stolz theorem:
$$ = lim_n to inftyfracsqrt2016^2 + 4n + sumZ(n+1)(n+1)sqrtn+1 - nsqrtn = 0 $$
Since the top degree of $n$ is $frac12$ and the bottom one is $ frac32$.
So can anyone tell me whether this solutions is correct? If it isn't, how would you solve it? If it is correct but my process is wrong I'd also like to know where I went wrong.
Thanks in advance.
real-analysis limits convergence
real-analysis limits convergence
asked 2 hours ago
Koy
1776
asked 2 hours ago
Koy
1776
asked 2 hours ago
Koy
1776
asked 2 hours ago
Koy
1776
asked 2 hours ago
Koy
1776
1776
add a comment |Â
add a comment |Â
1 Answer
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For $a$ we can go like this:
The sequence is clearly increasing, it's also positive, we just need to prove that the limit is infinite. If it were finite and equal to $ggeq a_1>0$, we would have
$$g=sqrtg^2+4+frac1g^222implies frac1g^222=-4 $$
which is impossible, so that $lim_ntoinfty a_n = infty$.
Your solution for $b$ looks good.
For $c$:
$$lim_ntoinfty fracsumlimits_k=1^n a_kn^3/2 = lim_ntoinfty fraca_n+1sqrtn+1+fracnsqrtn+1+sqrtn = lim_ntoinfty fraca_n(3/2)sqrtn = frac23cdot2 = frac43 $$
First equality is Stolz's theorem, $$(n+1)^3/2-n^3/2 = n(n+1)^1/2+(n+1)^1/2-n^3/2=\=n(sqrtn+1-sqrtn)+sqrtn+1=fracnsqrtn+1+sqrtn+sqrtn+1,$$ for the second we used that $fracfracnsqrtn+1+sqrtn+sqrtn+1sqrtn+1to 3/2$ as $ntoinfty$, third we calculated earlier.
answered 2 hours ago
Jakobian
1,963515
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
For $a$ we can go like this:
The sequence is clearly increasing, it's also positive, we just need to prove that the limit is infinite. If it were finite and equal to $ggeq a_1>0$, we would have
$$g=sqrtg^2+4+frac1g^222implies frac1g^222=-4 $$
which is impossible, so that $lim_ntoinfty a_n = infty$.
Your solution for $b$ looks good.
For $c$:
$$lim_ntoinfty fracsumlimits_k=1^n a_kn^3/2 = lim_ntoinfty fraca_n+1sqrtn+1+fracnsqrtn+1+sqrtn = lim_ntoinfty fraca_n(3/2)sqrtn = frac23cdot2 = frac43 $$
First equality is Stolz's theorem, $$(n+1)^3/2-n^3/2 = n(n+1)^1/2+(n+1)^1/2-n^3/2=\=n(sqrtn+1-sqrtn)+sqrtn+1=fracnsqrtn+1+sqrtn+sqrtn+1,$$ for the second we used that $fracfracnsqrtn+1+sqrtn+sqrtn+1sqrtn+1to 3/2$ as $ntoinfty$, third we calculated earlier.
answered 2 hours ago
Jakobian
1,963515
add a comment |Â
up vote
4
down vote
accepted
For $a$ we can go like this:
The sequence is clearly increasing, it's also positive, we just need to prove that the limit is infinite. If it were finite and equal to $ggeq a_1>0$, we would have
$$g=sqrtg^2+4+frac1g^222implies frac1g^222=-4 $$
which is impossible, so that $lim_ntoinfty a_n = infty$.
Your solution for $b$ looks good.
For $c$:
$$lim_ntoinfty fracsumlimits_k=1^n a_kn^3/2 = lim_ntoinfty fraca_n+1sqrtn+1+fracnsqrtn+1+sqrtn = lim_ntoinfty fraca_n(3/2)sqrtn = frac23cdot2 = frac43 $$
First equality is Stolz's theorem, $$(n+1)^3/2-n^3/2 = n(n+1)^1/2+(n+1)^1/2-n^3/2=\=n(sqrtn+1-sqrtn)+sqrtn+1=fracnsqrtn+1+sqrtn+sqrtn+1,$$ for the second we used that $fracfracnsqrtn+1+sqrtn+sqrtn+1sqrtn+1to 3/2$ as $ntoinfty$, third we calculated earlier.
answered 2 hours ago
Jakobian
1,963515
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
For $a$ we can go like this:
The sequence is clearly increasing, it's also positive, we just need to prove that the limit is infinite. If it were finite and equal to $ggeq a_1>0$, we would have
$$g=sqrtg^2+4+frac1g^222implies frac1g^222=-4 $$
which is impossible, so that $lim_ntoinfty a_n = infty$.
Your solution for $b$ looks good.
For $c$:
$$lim_ntoinfty fracsumlimits_k=1^n a_kn^3/2 = lim_ntoinfty fraca_n+1sqrtn+1+fracnsqrtn+1+sqrtn = lim_ntoinfty fraca_n(3/2)sqrtn = frac23cdot2 = frac43 $$
First equality is Stolz's theorem, $$(n+1)^3/2-n^3/2 = n(n+1)^1/2+(n+1)^1/2-n^3/2=\=n(sqrtn+1-sqrtn)+sqrtn+1=fracnsqrtn+1+sqrtn+sqrtn+1,$$ for the second we used that $fracfracnsqrtn+1+sqrtn+sqrtn+1sqrtn+1to 3/2$ as $ntoinfty$, third we calculated earlier.
answered 2 hours ago
Jakobian
1,963515
For $a$ we can go like this:
The sequence is clearly increasing, it's also positive, we just need to prove that the limit is infinite. If it were finite and equal to $ggeq a_1>0$, we would have
$$g=sqrtg^2+4+frac1g^222implies frac1g^222=-4 $$
which is impossible, so that $lim_ntoinfty a_n = infty$.
Your solution for $b$ looks good.
For $c$:
$$lim_ntoinfty fracsumlimits_k=1^n a_kn^3/2 = lim_ntoinfty fraca_n+1sqrtn+1+fracnsqrtn+1+sqrtn = lim_ntoinfty fraca_n(3/2)sqrtn = frac23cdot2 = frac43 $$
First equality is Stolz's theorem, $$(n+1)^3/2-n^3/2 = n(n+1)^1/2+(n+1)^1/2-n^3/2=\=n(sqrtn+1-sqrtn)+sqrtn+1=fracnsqrtn+1+sqrtn+sqrtn+1,$$ for the second we used that $fracfracnsqrtn+1+sqrtn+sqrtn+1sqrtn+1to 3/2$ as $ntoinfty$, third we calculated earlier.
answered 2 hours ago
Jakobian
1,963515
edited 2 hours ago
answered 2 hours ago
Jakobian
1,963515
answered 2 hours ago
Jakobian
1,963515
answered 2 hours ago
Jakobian
1,963515
1,963515
add a comment |Â
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