Limit of $a_n$ is -$2$ then consider $sum a_n ^{-n}$
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If the limit of $a_n$ is -$2$ then consider the infinite series $sum a_n ^{-n}$. Does this converge or diverge?
I thought that because of the exponent, the root test will be a good approach. Consider:
$$ lim_{nrightarrow infty} sqrt[n]{a_n^{-n}}=lim_{nrightarrow infty} frac{1}{a_n}= -frac{1}{2}<1$$ so the limit converges. Am I correct in saying this?
Edit: a condittion for the root test is that $a_n>0$ this is clearly not the case, my proof fails.
real-analysis proof-verification
asked Dec 1 '18 at 13:55
Wesley StrikWesley Strik
2,017423
$endgroup$
add a comment |
$begingroup$
If the limit of $a_n$ is -$2$ then consider the infinite series $sum a_n ^{-n}$. Does this converge or diverge?
I thought that because of the exponent, the root test will be a good approach. Consider:
$$ lim_{nrightarrow infty} sqrt[n]{a_n^{-n}}=lim_{nrightarrow infty} frac{1}{a_n}= -frac{1}{2}<1$$ so the limit converges. Am I correct in saying this?
Edit: a condittion for the root test is that $a_n>0$ this is clearly not the case, my proof fails.
real-analysis proof-verification
asked Dec 1 '18 at 13:55
Wesley StrikWesley Strik
2,017423
$endgroup$
1
$begingroup$
Not quite. Be careful about signs.
$endgroup$
– user10354138
Dec 1 '18 at 13:57
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oh yeah, a condition on the root test is that $a_n >0$
$endgroup$
– Wesley Strik
Dec 1 '18 at 13:58
2
$begingroup$
Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
$endgroup$
– Szeto
Dec 1 '18 at 14:02
$begingroup$
Or the alternating series test.
$endgroup$
– Arthur
Dec 1 '18 at 14:49
add a comment |
$begingroup$
If the limit of $a_n$ is -$2$ then consider the infinite series $sum a_n ^{-n}$. Does this converge or diverge?
I thought that because of the exponent, the root test will be a good approach. Consider:
$$ lim_{nrightarrow infty} sqrt[n]{a_n^{-n}}=lim_{nrightarrow infty} frac{1}{a_n}= -frac{1}{2}<1$$ so the limit converges. Am I correct in saying this?
Edit: a condittion for the root test is that $a_n>0$ this is clearly not the case, my proof fails.
real-analysis proof-verification
asked Dec 1 '18 at 13:55
Wesley StrikWesley Strik
2,017423
$endgroup$
If the limit of $a_n$ is -$2$ then consider the infinite series $sum a_n ^{-n}$. Does this converge or diverge?
I thought that because of the exponent, the root test will be a good approach. Consider:
$$ lim_{nrightarrow infty} sqrt[n]{a_n^{-n}}=lim_{nrightarrow infty} frac{1}{a_n}= -frac{1}{2}<1$$ so the limit converges. Am I correct in saying this?
Edit: a condittion for the root test is that $a_n>0$ this is clearly not the case, my proof fails.
real-analysis proof-verification
real-analysis proof-verification
asked Dec 1 '18 at 13:55
Wesley StrikWesley Strik
2,017423
asked Dec 1 '18 at 13:55
Wesley StrikWesley Strik
2,017423
edited Dec 1 '18 at 13:59
Wesley Strik
asked Dec 1 '18 at 13:55
Wesley StrikWesley Strik
2,017423
asked Dec 1 '18 at 13:55
Wesley StrikWesley Strik
2,017423
asked Dec 1 '18 at 13:55
Wesley StrikWesley Strik
2,017423
2,017423
1
$begingroup$
Not quite. Be careful about signs.
$endgroup$
– user10354138
Dec 1 '18 at 13:57
$begingroup$
oh yeah, a condition on the root test is that $a_n >0$
$endgroup$
– Wesley Strik
Dec 1 '18 at 13:58
2
$begingroup$
Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
$endgroup$
– Szeto
Dec 1 '18 at 14:02
$begingroup$
Or the alternating series test.
$endgroup$
– Arthur
Dec 1 '18 at 14:49
add a comment |
1
$begingroup$
Not quite. Be careful about signs.
$endgroup$
– user10354138
Dec 1 '18 at 13:57
$begingroup$
oh yeah, a condition on the root test is that $a_n >0$
$endgroup$
– Wesley Strik
Dec 1 '18 at 13:58
2
$begingroup$
Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
$endgroup$
– Szeto
Dec 1 '18 at 14:02
$begingroup$
Or the alternating series test.
$endgroup$
– Arthur
Dec 1 '18 at 14:49
1
1
$begingroup$
Not quite. Be careful about signs.
$endgroup$
– user10354138
Dec 1 '18 at 13:57
$begingroup$
Not quite. Be careful about signs.
$endgroup$
– user10354138
Dec 1 '18 at 13:57
$begingroup$
oh yeah, a condition on the root test is that $a_n >0$
$endgroup$
– Wesley Strik
Dec 1 '18 at 13:58
$begingroup$
oh yeah, a condition on the root test is that $a_n >0$
$endgroup$
– Wesley Strik
Dec 1 '18 at 13:58
2
2
$begingroup$
Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
$endgroup$
– Szeto
Dec 1 '18 at 14:02
$begingroup$
Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
$endgroup$
– Szeto
Dec 1 '18 at 14:02
$begingroup$
Or the alternating series test.
$endgroup$
– Arthur
Dec 1 '18 at 14:49
$begingroup$
Or the alternating series test.
$endgroup$
– Arthur
Dec 1 '18 at 14:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have that
$$a_nsim left(-frac{1}{2}right)^n$$
which clearly converges by geometric series.
As an alternative consider the $sum |a_n|$ and apply root test if you want use that.
answered Dec 1 '18 at 14:01
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Oh yeah is converges absolutely, so it converges.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:25
1
$begingroup$
@WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
$endgroup$
– gimusi
Dec 1 '18 at 14:36
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
We have that
$$a_nsim left(-frac{1}{2}right)^n$$
which clearly converges by geometric series.
As an alternative consider the $sum |a_n|$ and apply root test if you want use that.
answered Dec 1 '18 at 14:01
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Oh yeah is converges absolutely, so it converges.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:25
1
$begingroup$
@WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
$endgroup$
– gimusi
Dec 1 '18 at 14:36
add a comment |
$begingroup$
We have that
$$a_nsim left(-frac{1}{2}right)^n$$
which clearly converges by geometric series.
As an alternative consider the $sum |a_n|$ and apply root test if you want use that.
answered Dec 1 '18 at 14:01
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Oh yeah is converges absolutely, so it converges.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:25
1
$begingroup$
@WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
$endgroup$
– gimusi
Dec 1 '18 at 14:36
add a comment |
$begingroup$
We have that
$$a_nsim left(-frac{1}{2}right)^n$$
which clearly converges by geometric series.
As an alternative consider the $sum |a_n|$ and apply root test if you want use that.
answered Dec 1 '18 at 14:01
gimusigimusi
92.8k84494
$endgroup$
We have that
$$a_nsim left(-frac{1}{2}right)^n$$
which clearly converges by geometric series.
As an alternative consider the $sum |a_n|$ and apply root test if you want use that.
answered Dec 1 '18 at 14:01
gimusigimusi
92.8k84494
answered Dec 1 '18 at 14:01
gimusigimusi
92.8k84494
answered Dec 1 '18 at 14:01
gimusigimusi
92.8k84494
answered Dec 1 '18 at 14:01
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
Oh yeah is converges absolutely, so it converges.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:25
1
$begingroup$
@WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
$endgroup$
– gimusi
Dec 1 '18 at 14:36
add a comment |
$begingroup$
Oh yeah is converges absolutely, so it converges.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:25
1
$begingroup$
@WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
$endgroup$
– gimusi
Dec 1 '18 at 14:36
$begingroup$
Oh yeah is converges absolutely, so it converges.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:25
$begingroup$
Oh yeah is converges absolutely, so it converges.
$endgroup$
– Wesley Strik
Dec 1 '18 at 14:25
1
1
$begingroup$
@WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
$endgroup$
– gimusi
Dec 1 '18 at 14:36
$begingroup$
@WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
$endgroup$
– gimusi
Dec 1 '18 at 14:36
add a comment |
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1
$begingroup$
Not quite. Be careful about signs.
$endgroup$
– user10354138
Dec 1 '18 at 13:57
$begingroup$
oh yeah, a condition on the root test is that $a_n >0$
$endgroup$
– Wesley Strik
Dec 1 '18 at 13:58
2
$begingroup$
Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
$endgroup$
– Szeto
Dec 1 '18 at 14:02
$begingroup$
Or the alternating series test.
$endgroup$
– Arthur
Dec 1 '18 at 14:49