Determine whether the sequence converges or diverges.
$begingroup$
Let : $\a_n=frac{(-pi)^n}{4^n}$
What my teacher told me :
$a_1=frac{-pi}{4} approx -0.785\
a_2=frac{pi^2}{16} approx 0.617\
a_3=frac{-pi^3}{64} approx -0.484\
a_4=frac{pi^4}{256} approx 0.380\$
So the sequence diverges. But I'm not really sure about the answer.
Here is my teacher's work:
calculus sequences-and-series
New contributor
$endgroup$
add a comment |
$begingroup$
Let : $\a_n=frac{(-pi)^n}{4^n}$
What my teacher told me :
$a_1=frac{-pi}{4} approx -0.785\
a_2=frac{pi^2}{16} approx 0.617\
a_3=frac{-pi^3}{64} approx -0.484\
a_4=frac{pi^4}{256} approx 0.380\$
So the sequence diverges. But I'm not really sure about the answer.
Here is my teacher's work:
calculus sequences-and-series
New contributor
$endgroup$
$begingroup$
$a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
$endgroup$
– Clclstdnt
9 hours ago
$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
9 hours ago
1
$begingroup$
One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
$endgroup$
– user587192
9 hours ago
$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
9 hours ago
$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
9 hours ago
add a comment |
$begingroup$
Let : $\a_n=frac{(-pi)^n}{4^n}$
What my teacher told me :
$a_1=frac{-pi}{4} approx -0.785\
a_2=frac{pi^2}{16} approx 0.617\
a_3=frac{-pi^3}{64} approx -0.484\
a_4=frac{pi^4}{256} approx 0.380\$
So the sequence diverges. But I'm not really sure about the answer.
Here is my teacher's work:
calculus sequences-and-series
New contributor
$endgroup$
Let : $\a_n=frac{(-pi)^n}{4^n}$
What my teacher told me :
$a_1=frac{-pi}{4} approx -0.785\
a_2=frac{pi^2}{16} approx 0.617\
a_3=frac{-pi^3}{64} approx -0.484\
a_4=frac{pi^4}{256} approx 0.380\$
So the sequence diverges. But I'm not really sure about the answer.
Here is my teacher's work:
calculus sequences-and-series
calculus sequences-and-series
New contributor
New contributor
edited 8 hours ago
user587192
1,779315
1,779315
New contributor
asked 9 hours ago
airlanggaairlangga
72
72
New contributor
New contributor
$begingroup$
$a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
$endgroup$
– Clclstdnt
9 hours ago
$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
9 hours ago
1
$begingroup$
One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
$endgroup$
– user587192
9 hours ago
$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
9 hours ago
$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
9 hours ago
add a comment |
$begingroup$
$a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
$endgroup$
– Clclstdnt
9 hours ago
$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
9 hours ago
1
$begingroup$
One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
$endgroup$
– user587192
9 hours ago
$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
9 hours ago
$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
9 hours ago
$begingroup$
$a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
$endgroup$
– Clclstdnt
9 hours ago
$begingroup$
$a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
$endgroup$
– Clclstdnt
9 hours ago
$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
9 hours ago
$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
9 hours ago
1
1
$begingroup$
One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
$endgroup$
– user587192
9 hours ago
$begingroup$
One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
$endgroup$
– user587192
9 hours ago
$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
9 hours ago
$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
9 hours ago
$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
9 hours ago
$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
9 hours ago
add a comment |
2 Answers
2
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$begingroup$
By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that
$$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$
$endgroup$
add a comment |
$begingroup$
Well: $$|t|<1to lim_{ntoinfty}t^n=0$$
$endgroup$
add a comment |
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2 Answers
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$begingroup$
By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that
$$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$
$endgroup$
add a comment |
$begingroup$
By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that
$$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$
$endgroup$
add a comment |
$begingroup$
By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that
$$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$
$endgroup$
By triangle inequality one has that if $lim_{n to infty} |a_n| = 0$, then $lim_{n to infty} a_n = 0$. Thus, if $a_n = frac{(- pi )^n }{4^n } = frac{ (-1)^n pi^n }{4^n} $. Observe that
$$ left| frac{ (-1)^n pi^n }{4^n} right| = frac{ pi^n }{4^n} = left( frac{ pi }{4} right)^n to 0 $$
answered 9 hours ago
Jimmy SabaterJimmy Sabater
2,136219
2,136219
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$begingroup$
Well: $$|t|<1to lim_{ntoinfty}t^n=0$$
$endgroup$
add a comment |
$begingroup$
Well: $$|t|<1to lim_{ntoinfty}t^n=0$$
$endgroup$
add a comment |
$begingroup$
Well: $$|t|<1to lim_{ntoinfty}t^n=0$$
$endgroup$
Well: $$|t|<1to lim_{ntoinfty}t^n=0$$
answered 9 hours ago
Rhys HughesRhys Hughes
5,1901427
5,1901427
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$begingroup$
$a_n = (-1)^n left(frac{pi}{4}right)^n$ and $ frac{pi}{4}<1$ and this is an alternating series..... therefore by the Alternating Series Test this series.....
$endgroup$
– Clclstdnt
9 hours ago
$begingroup$
@Clclstdnt: this is a sequence, not a series. No alternating series test.
$endgroup$
– user587192
9 hours ago
1
$begingroup$
One would not simply list four terms and claim the sequence is divergent or convergent. You may have probably taken a wrong note on what your teacher said.
$endgroup$
– user587192
9 hours ago
$begingroup$
@user587192 I just attached her work
$endgroup$
– airlangga
9 hours ago
$begingroup$
@airlangga: Your teacher's point is that $frac{pi}{4}<1$. And the sequence converges.
$endgroup$
– user587192
9 hours ago