Mixing
Verification of the approach on finding the limit of $x_n$ as $ntoinfty$ given $x_{n+1} = sqrt[k]{5x_n}$...
$begingroup$
Let $x_n$ denote a sequence, $ninBbb N$:
$$
begin{cases}
x_{n+1} = sqrt[k]{5x_n}\
x_1 = sqrt[k]{5}\
kinBbb N
end{cases}
$$
Find:
$$
lim_{ntoinfty} sqrt[k]{x_n}
$$
I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
$$
x_1 = sqrt[k]{5}\
x_2 = sqrt[k]{5sqrt[k]{5}}\
x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
dots
$$
Rewrite as:
$$
begin{align}
x_1 &= 5^{1over k}\
x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
&cdots\
x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
end{align}
$$
This is hardly readable using powers in mathjax. Rewrite:
$$
log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
$$
Nominator is a regular geometric series:
$$
1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
$$
Thus:
$$
frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
$$
Since $a^x$ is continuous we may consider the limit of $(1)$:
$$
lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
$$
Which implies:
$$
lim_{ntoinfty}x_n = sqrt[k-1]{5}
$$
Is it the right way to approach this problem? Could it be simplified? Thank you!
calculus limits proof-verification
asked Jan 4 at 11:26
romanroman
1,98321221
$endgroup$
add a comment |
$begingroup$
Let $x_n$ denote a sequence, $ninBbb N$:
$$
begin{cases}
x_{n+1} = sqrt[k]{5x_n}\
x_1 = sqrt[k]{5}\
kinBbb N
end{cases}
$$
Find:
$$
lim_{ntoinfty} sqrt[k]{x_n}
$$
I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
$$
x_1 = sqrt[k]{5}\
x_2 = sqrt[k]{5sqrt[k]{5}}\
x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
dots
$$
Rewrite as:
$$
begin{align}
x_1 &= 5^{1over k}\
x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
&cdots\
x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
end{align}
$$
This is hardly readable using powers in mathjax. Rewrite:
$$
log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
$$
Nominator is a regular geometric series:
$$
1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
$$
Thus:
$$
frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
$$
Since $a^x$ is continuous we may consider the limit of $(1)$:
$$
lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
$$
Which implies:
$$
lim_{ntoinfty}x_n = sqrt[k-1]{5}
$$
Is it the right way to approach this problem? Could it be simplified? Thank you!
calculus limits proof-verification
asked Jan 4 at 11:26
romanroman
1,98321221
$endgroup$
add a comment |
$begingroup$
Let $x_n$ denote a sequence, $ninBbb N$:
$$
begin{cases}
x_{n+1} = sqrt[k]{5x_n}\
x_1 = sqrt[k]{5}\
kinBbb N
end{cases}
$$
Find:
$$
lim_{ntoinfty} sqrt[k]{x_n}
$$
I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
$$
x_1 = sqrt[k]{5}\
x_2 = sqrt[k]{5sqrt[k]{5}}\
x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
dots
$$
Rewrite as:
$$
begin{align}
x_1 &= 5^{1over k}\
x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
&cdots\
x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
end{align}
$$
This is hardly readable using powers in mathjax. Rewrite:
$$
log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
$$
Nominator is a regular geometric series:
$$
1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
$$
Thus:
$$
frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
$$
Since $a^x$ is continuous we may consider the limit of $(1)$:
$$
lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
$$
Which implies:
$$
lim_{ntoinfty}x_n = sqrt[k-1]{5}
$$
Is it the right way to approach this problem? Could it be simplified? Thank you!
calculus limits proof-verification
asked Jan 4 at 11:26
romanroman
1,98321221
$endgroup$
Let $x_n$ denote a sequence, $ninBbb N$:
$$
begin{cases}
x_{n+1} = sqrt[k]{5x_n}\
x_1 = sqrt[k]{5}\
kinBbb N
end{cases}
$$
Find:
$$
lim_{ntoinfty} sqrt[k]{x_n}
$$
I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms:
$$
x_1 = sqrt[k]{5}\
x_2 = sqrt[k]{5sqrt[k]{5}}\
x_3 = sqrt[k]{5sqrt[k]{5sqrt[k]{5}}}\
dots
$$
Rewrite as:
$$
begin{align}
x_1 &= 5^{1over k}\
x_2 &= left(5cdot 5^{1over k}right)^{1over k} = 5^{frac{k+1}{k^2}}\
&cdots\
x_n &= 5^{frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}}
end{align}
$$
This is hardly readable using powers in mathjax. Rewrite:
$$
log_5 x_n = frac{k^{n-1} + k^{n-2} + cdots + 1}{k^n}
$$
Nominator is a regular geometric series:
$$
1 + k + cdots + k^{n-2} + k^{n-1} = frac{k^n - 1}{k - 1}
$$
Thus:
$$
frac{k^n - 1}{k - 1} cdot frac{1}{k^n} = frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}tag 1
$$
Since $a^x$ is continuous we may consider the limit of $(1)$:
$$
lim_{ntoinfty} left(frac{k^n}{(k - 1)k^n} - frac{1}{(k-1)k^n}right) = frac{1}{k-1}
$$
Which implies:
$$
lim_{ntoinfty}x_n = sqrt[k-1]{5}
$$
Is it the right way to approach this problem? Could it be simplified? Thank you!
calculus limits proof-verification
calculus limits proof-verification
asked Jan 4 at 11:26
romanroman
1,98321221
asked Jan 4 at 11:26
romanroman
1,98321221
asked Jan 4 at 11:26
romanroman
1,98321221
asked Jan 4 at 11:26
romanroman
1,98321221
asked Jan 4 at 11:26
romanroman
1,98321221
1,98321221
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
$sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.
We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.
The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.
Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
$$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$
It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.
answered Jan 4 at 11:40
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
Thank you! Induction is indeed a better suit here.
$endgroup$
– roman
Jan 4 at 11:55
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061547%2fverification-of-the-approach-on-finding-the-limit-of-x-n-as-n-to-infty-given%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
$sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.
We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.
The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.
Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
$$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$
It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.
answered Jan 4 at 11:40
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
Thank you! Induction is indeed a better suit here.
$endgroup$
– roman
Jan 4 at 11:55
add a comment |
$begingroup$
We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
$sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.
We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.
The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.
Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
$$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$
It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.
answered Jan 4 at 11:40
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
Thank you! Induction is indeed a better suit here.
$endgroup$
– roman
Jan 4 at 11:55
add a comment |
$begingroup$
We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
$sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.
We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.
The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.
Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
$$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$
It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.
answered Jan 4 at 11:40
mechanodroidmechanodroid
27.1k62446
$endgroup$
We have $x_2 = sqrt[k]{5sqrt[k]{5}} ge sqrt[k]{5} = x_1$. If we assume that $x_{n} ge x_{n-1}$, it follows
$sqrt[k]{5x_{n}} ge sqrt[k]{5x_{n-1}}$, or $x_{n+1} ge x_n$. Induction implies that $(x_n)_n$ is increasing.
We have $x_1 = sqrt[k]{5} le sqrt[k-1]{5}$. If we assume $x_n le sqrt[k-1]{5}$, we get $$x_{n+1} = sqrt[k]{5x_n} le sqrt[k]{5sqrt[k-1]{5}} = left(5cdot 5^{frac1{k-1}}right)^{1/k} = sqrt[k-1]{5}$$
Induction implies that $x_n le sqrt[k-1]{5}, forall ninmathbb{N}$.
The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = lim_{ntoinfty} x_n$.
Letting $ntoinfty$ in the relation $x_{n+1} = sqrt[k]{5x_n}$ gives
$$L = sqrt[k]{5L} implies L(L^{k-1}-5) = 0$$
It cannot be $L = 0$ because $x_n ge x_1 =sqrt[k]{5}, forall ninmathbb{N}$. Therefore $L = sqrt[k-1]{5}$.
answered Jan 4 at 11:40
mechanodroidmechanodroid
27.1k62446
answered Jan 4 at 11:40
mechanodroidmechanodroid
27.1k62446
answered Jan 4 at 11:40
mechanodroidmechanodroid
27.1k62446
answered Jan 4 at 11:40
mechanodroidmechanodroid
27.1k62446
27.1k62446
$begingroup$
Thank you! Induction is indeed a better suit here.
$endgroup$
– roman
Jan 4 at 11:55
add a comment |
$begingroup$
Thank you! Induction is indeed a better suit here.
$endgroup$
– roman
Jan 4 at 11:55
$begingroup$
Thank you! Induction is indeed a better suit here.
$endgroup$
– roman
Jan 4 at 11:55
$begingroup$
Thank you! Induction is indeed a better suit here.
$endgroup$
– roman
Jan 4 at 11:55
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061547%2fverification-of-the-approach-on-finding-the-limit-of-x-n-as-n-to-infty-given%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
