Evaluate the limit $lim_{ntoinfty}log_aleft(frac{4^nn!}{n^n}right)$












3












$begingroup$



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?










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$endgroup$












  • $begingroup$
    Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
    $endgroup$
    – aleden
    Jan 7 at 18:22


















3












$begingroup$



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?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
    $endgroup$
    – aleden
    Jan 7 at 18:22
















3












3








3


1



$begingroup$



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?










share|cite|improve this question









$endgroup$





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






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asked Jan 7 at 17:49









romanroman

2,28421224




2,28421224












  • $begingroup$
    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






$begingroup$
Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
$endgroup$
– aleden
Jan 7 at 18:22












1 Answer
1






active

oldest

votes


















3












$begingroup$

Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












$begingroup$

Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$






share|cite|improve this answer











$endgroup$



Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 13:35

























answered Jan 7 at 17:59









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

75.6k42866




75.6k42866












  • $begingroup$
    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




$begingroup$
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


















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