sequence of $a^n$
$begingroup$
I am trying to show that when a $in Bbb R$ with $lvert arvert < 1$ the limit $lim_{nto+infty} a^n = 0$ using the definition of convergence of a sequence. So far I have
$lvert a^n - 0rvert < epsilon$
$lvert a^n rvert < epsilon$
$lvert arvert^n < epsilon$
and
$lvert a rvert < 1 $ $implies$ $lvert arvert^n < 1$
But I don't see where I am supposed to head with this in order to prove it? any advice would be greatly appreciated in understanding this. Thank you.
sequences-and-series analysis
edited Nov 29 '18 at 14:10
Tianlalu
3,08621038
asked Nov 29 '18 at 13:45
adsy9adsy9
32
$endgroup$
add a comment |
$begingroup$
I am trying to show that when a $in Bbb R$ with $lvert arvert < 1$ the limit $lim_{nto+infty} a^n = 0$ using the definition of convergence of a sequence. So far I have
$lvert a^n - 0rvert < epsilon$
$lvert a^n rvert < epsilon$
$lvert arvert^n < epsilon$
and
$lvert a rvert < 1 $ $implies$ $lvert arvert^n < 1$
But I don't see where I am supposed to head with this in order to prove it? any advice would be greatly appreciated in understanding this. Thank you.
sequences-and-series analysis
edited Nov 29 '18 at 14:10
Tianlalu
3,08621038
asked Nov 29 '18 at 13:45
adsy9adsy9
32
$endgroup$
$begingroup$
For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
$endgroup$
– Stockfish
Nov 29 '18 at 13:47
add a comment |
$begingroup$
I am trying to show that when a $in Bbb R$ with $lvert arvert < 1$ the limit $lim_{nto+infty} a^n = 0$ using the definition of convergence of a sequence. So far I have
$lvert a^n - 0rvert < epsilon$
$lvert a^n rvert < epsilon$
$lvert arvert^n < epsilon$
and
$lvert a rvert < 1 $ $implies$ $lvert arvert^n < 1$
But I don't see where I am supposed to head with this in order to prove it? any advice would be greatly appreciated in understanding this. Thank you.
sequences-and-series analysis
edited Nov 29 '18 at 14:10
Tianlalu
3,08621038
asked Nov 29 '18 at 13:45
adsy9adsy9
32
$endgroup$
I am trying to show that when a $in Bbb R$ with $lvert arvert < 1$ the limit $lim_{nto+infty} a^n = 0$ using the definition of convergence of a sequence. So far I have
$lvert a^n - 0rvert < epsilon$
$lvert a^n rvert < epsilon$
$lvert arvert^n < epsilon$
and
$lvert a rvert < 1 $ $implies$ $lvert arvert^n < 1$
But I don't see where I am supposed to head with this in order to prove it? any advice would be greatly appreciated in understanding this. Thank you.
sequences-and-series analysis
sequences-and-series analysis
edited Nov 29 '18 at 14:10
Tianlalu
3,08621038
asked Nov 29 '18 at 13:45
adsy9adsy9
32
edited Nov 29 '18 at 14:10
Tianlalu
3,08621038
asked Nov 29 '18 at 13:45
adsy9adsy9
32
edited Nov 29 '18 at 14:10
Tianlalu
3,08621038
edited Nov 29 '18 at 14:10
Tianlalu
3,08621038
edited Nov 29 '18 at 14:10
Tianlalu
3,08621038
3,08621038
asked Nov 29 '18 at 13:45
adsy9adsy9
32
asked Nov 29 '18 at 13:45
adsy9adsy9
32
asked Nov 29 '18 at 13:45
adsy9adsy9
32
32
$begingroup$
For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
$endgroup$
– Stockfish
Nov 29 '18 at 13:47
add a comment |
$begingroup$
For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
$endgroup$
– Stockfish
Nov 29 '18 at 13:47
$begingroup$
For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
$endgroup$
– Stockfish
Nov 29 '18 at 13:47
$begingroup$
For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
$endgroup$
– Stockfish
Nov 29 '18 at 13:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.
answered Nov 29 '18 at 13:49
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:
$frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$
Thus
$|a^n| < frac{1}{nh}$.
Can you proceed ?
answered Nov 29 '18 at 13:51
FredFred
45.7k1848
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
votes
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oldest
votes
$begingroup$
By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.
answered Nov 29 '18 at 13:49
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.
answered Nov 29 '18 at 13:49
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.
answered Nov 29 '18 at 13:49
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
By taking $ln$ from both sides we have $$nln |a|<ln epsilon$$and since $ln |a|<0$ (because $0<|a|<1$) we obtain $$n>{ln epsilonover ln |a|}$$therefore choose $N={ln epsilonover ln |a|}$ and you can conclude for any $n>N$ that $|a|^n<epsilon$ and the proof is complete.
answered Nov 29 '18 at 13:49
Mostafa AyazMostafa Ayaz
15.5k3939
answered Nov 29 '18 at 13:49
Mostafa AyazMostafa Ayaz
15.5k3939
answered Nov 29 '18 at 13:49
Mostafa AyazMostafa Ayaz
15.5k3939
answered Nov 29 '18 at 13:49
Mostafa AyazMostafa Ayaz
15.5k3939
15.5k3939
add a comment |
add a comment |
$begingroup$
If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:
$frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$
Thus
$|a^n| < frac{1}{nh}$.
Can you proceed ?
answered Nov 29 '18 at 13:51
FredFred
45.7k1848
$endgroup$
add a comment |
$begingroup$
If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:
$frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$
Thus
$|a^n| < frac{1}{nh}$.
Can you proceed ?
answered Nov 29 '18 at 13:51
FredFred
45.7k1848
$endgroup$
add a comment |
$begingroup$
If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:
$frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$
Thus
$|a^n| < frac{1}{nh}$.
Can you proceed ?
answered Nov 29 '18 at 13:51
FredFred
45.7k1848
$endgroup$
If $|a|<1$, then $1/|a|>1$, hence $frac{1}{|a|}=1+h$ with some $h>0$. Then we get, with Bernoulli's inequality:
$frac{1}{|a^n|}=(1+h)^n ge 1+nh > nh.$
Thus
$|a^n| < frac{1}{nh}$.
Can you proceed ?
answered Nov 29 '18 at 13:51
FredFred
45.7k1848
answered Nov 29 '18 at 13:51
FredFred
45.7k1848
answered Nov 29 '18 at 13:51
FredFred
45.7k1848
answered Nov 29 '18 at 13:51
FredFred
45.7k1848
45.7k1848
add a comment |
add a comment |
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$begingroup$
For given $varepsilon >0$, you have to find $N$ such that for every $n>N$ $|a|^n < varepsilon$ holds. Note that $N$ may depend on $a$. Note further that $n mapsto |a|^n$ is decreasing.
$endgroup$
– Stockfish
Nov 29 '18 at 13:47