Evaluate the limit $lim_{ntoinfty}log_aleft(frac{4^nn!}{n^n}right)$
$begingroup$
Evaluate the limit:
$$
lim_{ntoinfty}log_aleft(frac{4^nn!}{n^n}right)\
a>0\
a ne 1
$$
I've started with defining another sequence. Let:
$$
y_n = a^{x_n} = frac{4^nn!}{n^n}
$$
Consider the fraction:
$$
frac{y_{n+1}}{y_n} = frac{4^{n+1}(n+1)!}{(n+1)^{n+1}} cdot frac{n^n}{4^nn!}\
= frac{4n^n}{(n+1)^n}
$$
Consider the limit:
$$
begin{align}
lim_{ntoinfty}frac{y_{n+1}}{y_n} &= lim_{ntoinfty}frac{4n^n}{(n+1)^n} \
&= lim_{ntoinfty}4left(frac{n}{n+1} right)^n \
&= {4over e} > 1
end{align}
$$
So by this $y_n$ is divergent. Which means:
$$
lim_{ntoinfty}y_n = infty
$$
Now I'm having difficulties translating it in a backward direction. We have that:
$$
lim_{ntoinfty}y_n = lim_{ntoinfty}a^{x_n} = infty
$$
Or:
$$
log_a lim_{ntoinfty}a^{x_n} = log_a(infty)
$$
The answer suggests that:
$$
lim_{ntoinfty}x_n =
begin{cases}
+infty, a > 1\
-infty, 0 < a < 1
end{cases}
$$
And I don't see where this appears when going backward from $a^{x_n}$ to $x_n$. Could you please explain that to me?
calculus sequences-and-series limits
asked Jan 7 at 17:49
romanroman
2,41121225
$endgroup$
add a comment |
$begingroup$
Evaluate the limit:
$$
lim_{ntoinfty}log_aleft(frac{4^nn!}{n^n}right)\
a>0\
a ne 1
$$
I've started with defining another sequence. Let:
$$
y_n = a^{x_n} = frac{4^nn!}{n^n}
$$
Consider the fraction:
$$
frac{y_{n+1}}{y_n} = frac{4^{n+1}(n+1)!}{(n+1)^{n+1}} cdot frac{n^n}{4^nn!}\
= frac{4n^n}{(n+1)^n}
$$
Consider the limit:
$$
begin{align}
lim_{ntoinfty}frac{y_{n+1}}{y_n} &= lim_{ntoinfty}frac{4n^n}{(n+1)^n} \
&= lim_{ntoinfty}4left(frac{n}{n+1} right)^n \
&= {4over e} > 1
end{align}
$$
So by this $y_n$ is divergent. Which means:
$$
lim_{ntoinfty}y_n = infty
$$
Now I'm having difficulties translating it in a backward direction. We have that:
$$
lim_{ntoinfty}y_n = lim_{ntoinfty}a^{x_n} = infty
$$
Or:
$$
log_a lim_{ntoinfty}a^{x_n} = log_a(infty)
$$
The answer suggests that:
$$
lim_{ntoinfty}x_n =
begin{cases}
+infty, a > 1\
-infty, 0 < a < 1
end{cases}
$$
And I don't see where this appears when going backward from $a^{x_n}$ to $x_n$. Could you please explain that to me?
calculus sequences-and-series limits
asked Jan 7 at 17:49
romanroman
2,41121225
$endgroup$
$begingroup$
Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
$endgroup$
– aleden
Jan 7 at 18:22
add a comment |
$begingroup$
Evaluate the limit:
$$
lim_{ntoinfty}log_aleft(frac{4^nn!}{n^n}right)\
a>0\
a ne 1
$$
I've started with defining another sequence. Let:
$$
y_n = a^{x_n} = frac{4^nn!}{n^n}
$$
Consider the fraction:
$$
frac{y_{n+1}}{y_n} = frac{4^{n+1}(n+1)!}{(n+1)^{n+1}} cdot frac{n^n}{4^nn!}\
= frac{4n^n}{(n+1)^n}
$$
Consider the limit:
$$
begin{align}
lim_{ntoinfty}frac{y_{n+1}}{y_n} &= lim_{ntoinfty}frac{4n^n}{(n+1)^n} \
&= lim_{ntoinfty}4left(frac{n}{n+1} right)^n \
&= {4over e} > 1
end{align}
$$
So by this $y_n$ is divergent. Which means:
$$
lim_{ntoinfty}y_n = infty
$$
Now I'm having difficulties translating it in a backward direction. We have that:
$$
lim_{ntoinfty}y_n = lim_{ntoinfty}a^{x_n} = infty
$$
Or:
$$
log_a lim_{ntoinfty}a^{x_n} = log_a(infty)
$$
The answer suggests that:
$$
lim_{ntoinfty}x_n =
begin{cases}
+infty, a > 1\
-infty, 0 < a < 1
end{cases}
$$
And I don't see where this appears when going backward from $a^{x_n}$ to $x_n$. Could you please explain that to me?
calculus sequences-and-series limits
asked Jan 7 at 17:49
romanroman
2,41121225
$endgroup$
Evaluate the limit:
$$
lim_{ntoinfty}log_aleft(frac{4^nn!}{n^n}right)\
a>0\
a ne 1
$$
I've started with defining another sequence. Let:
$$
y_n = a^{x_n} = frac{4^nn!}{n^n}
$$
Consider the fraction:
$$
frac{y_{n+1}}{y_n} = frac{4^{n+1}(n+1)!}{(n+1)^{n+1}} cdot frac{n^n}{4^nn!}\
= frac{4n^n}{(n+1)^n}
$$
Consider the limit:
$$
begin{align}
lim_{ntoinfty}frac{y_{n+1}}{y_n} &= lim_{ntoinfty}frac{4n^n}{(n+1)^n} \
&= lim_{ntoinfty}4left(frac{n}{n+1} right)^n \
&= {4over e} > 1
end{align}
$$
So by this $y_n$ is divergent. Which means:
$$
lim_{ntoinfty}y_n = infty
$$
Now I'm having difficulties translating it in a backward direction. We have that:
$$
lim_{ntoinfty}y_n = lim_{ntoinfty}a^{x_n} = infty
$$
Or:
$$
log_a lim_{ntoinfty}a^{x_n} = log_a(infty)
$$
The answer suggests that:
$$
lim_{ntoinfty}x_n =
begin{cases}
+infty, a > 1\
-infty, 0 < a < 1
end{cases}
$$
And I don't see where this appears when going backward from $a^{x_n}$ to $x_n$. Could you please explain that to me?
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Jan 7 at 17:49
romanroman
2,41121225
asked Jan 7 at 17:49
romanroman
2,41121225
asked Jan 7 at 17:49
romanroman
2,41121225
asked Jan 7 at 17:49
romanroman
2,41121225
asked Jan 7 at 17:49
romanroman
2,41121225
2,41121225
$begingroup$
Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
$endgroup$
– aleden
Jan 7 at 18:22
add a comment |
$begingroup$
Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
$endgroup$
– aleden
Jan 7 at 18:22
$begingroup$
Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
$endgroup$
– aleden
Jan 7 at 18:22
$begingroup$
Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
$endgroup$
– aleden
Jan 7 at 18:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
$begingroup$
Well, I thought I understood what you meant, but then I returned back to my notes and realized thit is still unclear to me. I would greatly appreciate if you could elaborate on what follows from rewriting $log_a x$ as $frac{ln x}{ln a}$.
$endgroup$
– roman
Jan 9 at 14:37
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%2f3065275%2fevaluate-the-limit-lim-n-to-infty-log-a-left-frac4nnnn-right%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
$begingroup$
Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
$begingroup$
Well, I thought I understood what you meant, but then I returned back to my notes and realized thit is still unclear to me. I would greatly appreciate if you could elaborate on what follows from rewriting $log_a x$ as $frac{ln x}{ln a}$.
$endgroup$
– roman
Jan 9 at 14:37
add a comment |
$begingroup$
Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
$begingroup$
Well, I thought I understood what you meant, but then I returned back to my notes and realized thit is still unclear to me. I would greatly appreciate if you could elaborate on what follows from rewriting $log_a x$ as $frac{ln x}{ln a}$.
$endgroup$
– roman
Jan 9 at 14:37
add a comment |
$begingroup$
Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
$endgroup$
Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
edited Jan 8 at 13:35
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.2k42867
78.2k42867
$begingroup$
Well, I thought I understood what you meant, but then I returned back to my notes and realized thit is still unclear to me. I would greatly appreciate if you could elaborate on what follows from rewriting $log_a x$ as $frac{ln x}{ln a}$.
$endgroup$
– roman
Jan 9 at 14:37
add a comment |
$begingroup$
Well, I thought I understood what you meant, but then I returned back to my notes and realized thit is still unclear to me. I would greatly appreciate if you could elaborate on what follows from rewriting $log_a x$ as $frac{ln x}{ln a}$.
$endgroup$
– roman
Jan 9 at 14:37
$begingroup$
Well, I thought I understood what you meant, but then I returned back to my notes and realized thit is still unclear to me. I would greatly appreciate if you could elaborate on what follows from rewriting $log_a x$ as $frac{ln x}{ln a}$.
$endgroup$
– roman
Jan 9 at 14:37
$begingroup$
Well, I thought I understood what you meant, but then I returned back to my notes and realized thit is still unclear to me. I would greatly appreciate if you could elaborate on what follows from rewriting $log_a x$ as $frac{ln x}{ln a}$.
$endgroup$
– roman
Jan 9 at 14:37
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%2f3065275%2fevaluate-the-limit-lim-n-to-infty-log-a-left-frac4nnnn-right%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
$begingroup$
Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
$endgroup$
– aleden
Jan 7 at 18:22