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
2,28421224
$endgroup$
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
2,28421224
$endgroup$
$begingroup$
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
2,28421224
$endgroup$
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
2,28421224
asked Jan 7 at 17:49
romanroman
2,28421224
asked Jan 7 at 17:49
romanroman
2,28421224
asked Jan 7 at 17:49
romanroman
2,28421224
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
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
75.6k42866
$endgroup$
$begingroup$
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
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
$endgroup$
$begingroup$
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
75.6k42866
$endgroup$
$begingroup$
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
75.6k42866
$endgroup$
Hint: Note that $$log_a{x}=frac{ln(x)}{ln(a)}$$
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
edited Jan 8 at 13:35
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
answered Jan 7 at 17:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
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
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 |
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$begingroup$
Rewrite the expression in terms of the natural logarithm and then use Stirling's Approximation.
$endgroup$
– aleden
Jan 7 at 18:22