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.










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  • 1




    Not quite. Be careful about signs.
    – user10354138
    Dec 1 at 13:57










  • oh yeah, a condition on the root test is that $a_n >0$
    – WesleyGroupshaveFeelingsToo
    Dec 1 at 13:58






  • 2




    Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
    – Szeto
    Dec 1 at 14:02










  • Or the alternating series test.
    – Arthur
    Dec 1 at 14:49















up vote
0
down vote

favorite













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.










share|cite|improve this question




















  • 1




    Not quite. Be careful about signs.
    – user10354138
    Dec 1 at 13:57










  • oh yeah, a condition on the root test is that $a_n >0$
    – WesleyGroupshaveFeelingsToo
    Dec 1 at 13:58






  • 2




    Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
    – Szeto
    Dec 1 at 14:02










  • Or the alternating series test.
    – Arthur
    Dec 1 at 14:49













up vote
0
down vote

favorite









up vote
0
down vote

favorite












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.










share|cite|improve this question
















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






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share|cite|improve this question













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edited Dec 1 at 13:59

























asked Dec 1 at 13:55









WesleyGroupshaveFeelingsToo

1,163322




1,163322








  • 1




    Not quite. Be careful about signs.
    – user10354138
    Dec 1 at 13:57










  • oh yeah, a condition on the root test is that $a_n >0$
    – WesleyGroupshaveFeelingsToo
    Dec 1 at 13:58






  • 2




    Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
    – Szeto
    Dec 1 at 14:02










  • Or the alternating series test.
    – Arthur
    Dec 1 at 14:49














  • 1




    Not quite. Be careful about signs.
    – user10354138
    Dec 1 at 13:57










  • oh yeah, a condition on the root test is that $a_n >0$
    – WesleyGroupshaveFeelingsToo
    Dec 1 at 13:58






  • 2




    Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
    – Szeto
    Dec 1 at 14:02










  • Or the alternating series test.
    – Arthur
    Dec 1 at 14:49








1




1




Not quite. Be careful about signs.
– user10354138
Dec 1 at 13:57




Not quite. Be careful about signs.
– user10354138
Dec 1 at 13:57












oh yeah, a condition on the root test is that $a_n >0$
– WesleyGroupshaveFeelingsToo
Dec 1 at 13:58




oh yeah, a condition on the root test is that $a_n >0$
– WesleyGroupshaveFeelingsToo
Dec 1 at 13:58




2




2




Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
– Szeto
Dec 1 at 14:02




Consider $sum a_n^{-n}lesum |a_n|^{-n}$ and apply root test.
– Szeto
Dec 1 at 14:02












Or the alternating series test.
– Arthur
Dec 1 at 14:49




Or the alternating series test.
– Arthur
Dec 1 at 14:49










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










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.






share|cite|improve this answer





















  • Oh yeah is converges absolutely, so it converges.
    – WesleyGroupshaveFeelingsToo
    Dec 1 at 14:25






  • 1




    @WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
    – gimusi
    Dec 1 at 14:36











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










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.






share|cite|improve this answer





















  • Oh yeah is converges absolutely, so it converges.
    – WesleyGroupshaveFeelingsToo
    Dec 1 at 14:25






  • 1




    @WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
    – gimusi
    Dec 1 at 14:36















up vote
1
down vote



accepted










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.






share|cite|improve this answer





















  • Oh yeah is converges absolutely, so it converges.
    – WesleyGroupshaveFeelingsToo
    Dec 1 at 14:25






  • 1




    @WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
    – gimusi
    Dec 1 at 14:36













up vote
1
down vote



accepted







up vote
1
down vote



accepted






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 at 14:01









gimusi

89.7k74495




89.7k74495












  • Oh yeah is converges absolutely, so it converges.
    – WesleyGroupshaveFeelingsToo
    Dec 1 at 14:25






  • 1




    @WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
    – gimusi
    Dec 1 at 14:36


















  • Oh yeah is converges absolutely, so it converges.
    – WesleyGroupshaveFeelingsToo
    Dec 1 at 14:25






  • 1




    @WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
    – gimusi
    Dec 1 at 14:36
















Oh yeah is converges absolutely, so it converges.
– WesleyGroupshaveFeelingsToo
Dec 1 at 14:25




Oh yeah is converges absolutely, so it converges.
– WesleyGroupshaveFeelingsToo
Dec 1 at 14:25




1




1




@WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
– gimusi
Dec 1 at 14:36




@WesleyGroupshaveFeelingsToo Yes and indeed I think that th eproper way to prove that is to consider the absolute convergence.
– gimusi
Dec 1 at 14:36


















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