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Convergence of the series $sum_n=1^infty frac(-1)^n-1x^n$.
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up vote
1
down vote
favorite
Show that $$sum_n=1^infty frac(-1)^n-1x^n$$ converges for every $x>1$.
let $a(x)$ be the sum of the series. does $a$ continious at $x=2$? differentiable?
I guess the first part is with leibniz but I am not sure about it.
calculus sequences-and-series convergence
edited Jul 16 at 19:14
Nosrati
19.8k41644
asked Jul 16 at 17:53
UltimateMath
385
add a comment |Â
up vote
1
down vote
favorite
Show that $$sum_n=1^infty frac(-1)^n-1x^n$$ converges for every $x>1$.
let $a(x)$ be the sum of the series. does $a$ continious at $x=2$? differentiable?
I guess the first part is with leibniz but I am not sure about it.
calculus sequences-and-series convergence
edited Jul 16 at 19:14
Nosrati
19.8k41644
asked Jul 16 at 17:53
UltimateMath
385
What convergence tests do you know? Since this summand is a simple ratio, there is an "obvious" choice for a test to try.
– Eric Towers
Jul 16 at 17:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that $$sum_n=1^infty frac(-1)^n-1x^n$$ converges for every $x>1$.
let $a(x)$ be the sum of the series. does $a$ continious at $x=2$? differentiable?
I guess the first part is with leibniz but I am not sure about it.
calculus sequences-and-series convergence
edited Jul 16 at 19:14
Nosrati
19.8k41644
asked Jul 16 at 17:53
UltimateMath
385
Show that $$sum_n=1^infty frac(-1)^n-1x^n$$ converges for every $x>1$.
let $a(x)$ be the sum of the series. does $a$ continious at $x=2$? differentiable?
I guess the first part is with leibniz but I am not sure about it.
calculus sequences-and-series convergence
edited Jul 16 at 19:14
Nosrati
19.8k41644
asked Jul 16 at 17:53
UltimateMath
385
edited Jul 16 at 19:14
Nosrati
19.8k41644
edited Jul 16 at 19:14
Nosrati
19.8k41644
edited Jul 16 at 19:14
Nosrati
19.8k41644
19.8k41644
asked Jul 16 at 17:53
UltimateMath
385
asked Jul 16 at 17:53
UltimateMath
385
asked Jul 16 at 17:53
UltimateMath
385
385
What convergence tests do you know? Since this summand is a simple ratio, there is an "obvious" choice for a test to try.
– Eric Towers
Jul 16 at 17:59
add a comment |Â
What convergence tests do you know? Since this summand is a simple ratio, there is an "obvious" choice for a test to try.
– Eric Towers
Jul 16 at 17:59
What convergence tests do you know? Since this summand is a simple ratio, there is an "obvious" choice for a test to try.
– Eric Towers
Jul 16 at 17:59
What convergence tests do you know? Since this summand is a simple ratio, there is an "obvious" choice for a test to try.
– Eric Towers
Jul 16 at 17:59
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
Hint:
Use the geometric series:
$$a(x) = frac1xsum_n=0^infty frac(-1)^nx^n = frac1xleft(1+frac1xright) = frac1x+1$$
answered Jul 16 at 17:59
mechanodroid
22.3k52041
why there is 1/x before the series?
– UltimateMath
Jul 16 at 18:31
@UltimateMath $$sum_n=1^infty frac(-1)^n-1x^n = frac1xsum_n=1^infty frac(-1)^n-1x^n-1 = frac1xsum_n=0^infty frac(-1)^nx^n$$
– mechanodroid
Jul 16 at 18:39
add a comment |Â
up vote
1
down vote
Using root test
$$lim_ntoinftysqrt[n]dfrac(-1)^nx^nright=dfrac1<1$$
then the series is converge for $|x|>1$.
answered Jul 16 at 18:00
Nosrati
19.8k41644
let a(x) be the sum of the series. does a continious at x=2? differentiable? any clue?
– UltimateMath
Jul 16 at 18:03
Clearly according to other answer, in $|x|>1$, the series converges to $dfrac1x+1$ which is continous and differentiable as well.
– Nosrati
Jul 16 at 18:05
add a comment |Â
up vote
0
down vote
Hint:
What if you use the ratio test?$$lim_ntoinfty|dfraca_n+1a_n|$$
answered Jul 16 at 18:03
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
0
down vote
Let's look at the
partial sums,
and let $y = -1/x$
so
$-1 < y < 0$..
$beginarray\
s_m(y)
&=sum_n=1^m (-1)^n-1(-y)^n\
&=sum_n=1^m (-1)^n-1(-1)^ny^n\
&=-sum_n=1^m y^n\
&=-ysum_n=0^m-1 y^n\
&=-ydfrac1-y^m1-y\
&=dfrac-y1-y-dfrac-y^m+11-y\
endarray
$
so
$beginarray\
s_m(y)+dfracy1-y
&=dfracy^m+11-y\
endarray
$
Therefore
$sum_n=1^m frac(-1)^n-1x^n+dfrac-1/x1+1/x
=dfrac1(-x)^m+1(1+1/x)
$
or
$sum_n=1^m frac(-1)^n-1x^n-dfrac1x+1
=dfrac(-1)^m+1x^m(x+1)
$.
What is needed now is
to show that
$lim_m to infty dfrac1x^m(x+1)
=0$.
(This is from
"What is Mathematics")
Since $x > 1$,
$x = 1+z$
where $z > 0$.
By Bernoulli's inequality,
$x^m
=(1+z)^m
ge 1+mz
gt mz
=m(x-1)$,
so
$ dfrac1x^m(x+1)
lt dfrac1m(x-1)(x+1)
=dfrac1m(x^2-1)
$,
so to make
$dfrac1x^m(x+1)
lt epsilon
$
it is enough to take
$m
gt dfrac1epsilon(x^2-1)
$.
This is certainly
not the best $m$,
but it is
completely elementary.
answered Jul 16 at 20:19
marty cohen
69.3k446122
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Hint:
Use the geometric series:
$$a(x) = frac1xsum_n=0^infty frac(-1)^nx^n = frac1xleft(1+frac1xright) = frac1x+1$$
answered Jul 16 at 17:59
mechanodroid
22.3k52041
why there is 1/x before the series?
– UltimateMath
Jul 16 at 18:31
@UltimateMath $$sum_n=1^infty frac(-1)^n-1x^n = frac1xsum_n=1^infty frac(-1)^n-1x^n-1 = frac1xsum_n=0^infty frac(-1)^nx^n$$
– mechanodroid
Jul 16 at 18:39
add a comment |Â
up vote
3
down vote
accepted
Hint:
Use the geometric series:
$$a(x) = frac1xsum_n=0^infty frac(-1)^nx^n = frac1xleft(1+frac1xright) = frac1x+1$$
answered Jul 16 at 17:59
mechanodroid
22.3k52041
why there is 1/x before the series?
– UltimateMath
Jul 16 at 18:31
@UltimateMath $$sum_n=1^infty frac(-1)^n-1x^n = frac1xsum_n=1^infty frac(-1)^n-1x^n-1 = frac1xsum_n=0^infty frac(-1)^nx^n$$
– mechanodroid
Jul 16 at 18:39
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Hint:
Use the geometric series:
$$a(x) = frac1xsum_n=0^infty frac(-1)^nx^n = frac1xleft(1+frac1xright) = frac1x+1$$
answered Jul 16 at 17:59
mechanodroid
22.3k52041
Hint:
Use the geometric series:
$$a(x) = frac1xsum_n=0^infty frac(-1)^nx^n = frac1xleft(1+frac1xright) = frac1x+1$$
answered Jul 16 at 17:59
mechanodroid
22.3k52041
answered Jul 16 at 17:59
mechanodroid
22.3k52041
answered Jul 16 at 17:59
mechanodroid
22.3k52041
answered Jul 16 at 17:59
mechanodroid
22.3k52041
22.3k52041
why there is 1/x before the series?
– UltimateMath
Jul 16 at 18:31
@UltimateMath $$sum_n=1^infty frac(-1)^n-1x^n = frac1xsum_n=1^infty frac(-1)^n-1x^n-1 = frac1xsum_n=0^infty frac(-1)^nx^n$$
– mechanodroid
Jul 16 at 18:39
add a comment |Â
why there is 1/x before the series?
– UltimateMath
Jul 16 at 18:31
@UltimateMath $$sum_n=1^infty frac(-1)^n-1x^n = frac1xsum_n=1^infty frac(-1)^n-1x^n-1 = frac1xsum_n=0^infty frac(-1)^nx^n$$
– mechanodroid
Jul 16 at 18:39
why there is 1/x before the series?
– UltimateMath
Jul 16 at 18:31
why there is 1/x before the series?
– UltimateMath
Jul 16 at 18:31
@UltimateMath $$sum_n=1^infty frac(-1)^n-1x^n = frac1xsum_n=1^infty frac(-1)^n-1x^n-1 = frac1xsum_n=0^infty frac(-1)^nx^n$$
– mechanodroid
Jul 16 at 18:39
@UltimateMath $$sum_n=1^infty frac(-1)^n-1x^n = frac1xsum_n=1^infty frac(-1)^n-1x^n-1 = frac1xsum_n=0^infty frac(-1)^nx^n$$
– mechanodroid
Jul 16 at 18:39
add a comment |Â
up vote
1
down vote
Using root test
$$lim_ntoinftysqrt[n]dfrac(-1)^nx^nright=dfrac1<1$$
then the series is converge for $|x|>1$.
answered Jul 16 at 18:00
Nosrati
19.8k41644
let a(x) be the sum of the series. does a continious at x=2? differentiable? any clue?
– UltimateMath
Jul 16 at 18:03
Clearly according to other answer, in $|x|>1$, the series converges to $dfrac1x+1$ which is continous and differentiable as well.
– Nosrati
Jul 16 at 18:05
add a comment |Â
up vote
1
down vote
Using root test
$$lim_ntoinftysqrt[n]dfrac(-1)^nx^nright=dfrac1<1$$
then the series is converge for $|x|>1$.
answered Jul 16 at 18:00
Nosrati
19.8k41644
let a(x) be the sum of the series. does a continious at x=2? differentiable? any clue?
– UltimateMath
Jul 16 at 18:03
Clearly according to other answer, in $|x|>1$, the series converges to $dfrac1x+1$ which is continous and differentiable as well.
– Nosrati
Jul 16 at 18:05
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Using root test
$$lim_ntoinftysqrt[n]dfrac(-1)^nx^nright=dfrac1<1$$
then the series is converge for $|x|>1$.
answered Jul 16 at 18:00
Nosrati
19.8k41644
Using root test
$$lim_ntoinftysqrt[n]dfrac(-1)^nx^nright=dfrac1<1$$
then the series is converge for $|x|>1$.
answered Jul 16 at 18:00
Nosrati
19.8k41644
answered Jul 16 at 18:00
Nosrati
19.8k41644
answered Jul 16 at 18:00
Nosrati
19.8k41644
answered Jul 16 at 18:00
Nosrati
19.8k41644
19.8k41644
let a(x) be the sum of the series. does a continious at x=2? differentiable? any clue?
– UltimateMath
Jul 16 at 18:03
Clearly according to other answer, in $|x|>1$, the series converges to $dfrac1x+1$ which is continous and differentiable as well.
– Nosrati
Jul 16 at 18:05
add a comment |Â
let a(x) be the sum of the series. does a continious at x=2? differentiable? any clue?
– UltimateMath
Jul 16 at 18:03
Clearly according to other answer, in $|x|>1$, the series converges to $dfrac1x+1$ which is continous and differentiable as well.
– Nosrati
Jul 16 at 18:05
let a(x) be the sum of the series. does a continious at x=2? differentiable? any clue?
– UltimateMath
Jul 16 at 18:03
let a(x) be the sum of the series. does a continious at x=2? differentiable? any clue?
– UltimateMath
Jul 16 at 18:03
Clearly according to other answer, in $|x|>1$, the series converges to $dfrac1x+1$ which is continous and differentiable as well.
– Nosrati
Jul 16 at 18:05
Clearly according to other answer, in $|x|>1$, the series converges to $dfrac1x+1$ which is continous and differentiable as well.
– Nosrati
Jul 16 at 18:05
add a comment |Â
up vote
0
down vote
Hint:
What if you use the ratio test?$$lim_ntoinfty|dfraca_n+1a_n|$$
answered Jul 16 at 18:03
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
0
down vote
Hint:
What if you use the ratio test?$$lim_ntoinfty|dfraca_n+1a_n|$$
answered Jul 16 at 18:03
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
What if you use the ratio test?$$lim_ntoinfty|dfraca_n+1a_n|$$
answered Jul 16 at 18:03
Mostafa Ayaz
8,6023630
Hint:
What if you use the ratio test?$$lim_ntoinfty|dfraca_n+1a_n|$$
answered Jul 16 at 18:03
Mostafa Ayaz
8,6023630
answered Jul 16 at 18:03
Mostafa Ayaz
8,6023630
answered Jul 16 at 18:03
Mostafa Ayaz
8,6023630
answered Jul 16 at 18:03
Mostafa Ayaz
8,6023630
8,6023630
add a comment |Â
add a comment |Â
up vote
0
down vote
Let's look at the
partial sums,
and let $y = -1/x$
so
$-1 < y < 0$..
$beginarray\
s_m(y)
&=sum_n=1^m (-1)^n-1(-y)^n\
&=sum_n=1^m (-1)^n-1(-1)^ny^n\
&=-sum_n=1^m y^n\
&=-ysum_n=0^m-1 y^n\
&=-ydfrac1-y^m1-y\
&=dfrac-y1-y-dfrac-y^m+11-y\
endarray
$
so
$beginarray\
s_m(y)+dfracy1-y
&=dfracy^m+11-y\
endarray
$
Therefore
$sum_n=1^m frac(-1)^n-1x^n+dfrac-1/x1+1/x
=dfrac1(-x)^m+1(1+1/x)
$
or
$sum_n=1^m frac(-1)^n-1x^n-dfrac1x+1
=dfrac(-1)^m+1x^m(x+1)
$.
What is needed now is
to show that
$lim_m to infty dfrac1x^m(x+1)
=0$.
(This is from
"What is Mathematics")
Since $x > 1$,
$x = 1+z$
where $z > 0$.
By Bernoulli's inequality,
$x^m
=(1+z)^m
ge 1+mz
gt mz
=m(x-1)$,
so
$ dfrac1x^m(x+1)
lt dfrac1m(x-1)(x+1)
=dfrac1m(x^2-1)
$,
so to make
$dfrac1x^m(x+1)
lt epsilon
$
it is enough to take
$m
gt dfrac1epsilon(x^2-1)
$.
This is certainly
not the best $m$,
but it is
completely elementary.
answered Jul 16 at 20:19
marty cohen
69.3k446122
add a comment |Â
up vote
0
down vote
Let's look at the
partial sums,
and let $y = -1/x$
so
$-1 < y < 0$..
$beginarray\
s_m(y)
&=sum_n=1^m (-1)^n-1(-y)^n\
&=sum_n=1^m (-1)^n-1(-1)^ny^n\
&=-sum_n=1^m y^n\
&=-ysum_n=0^m-1 y^n\
&=-ydfrac1-y^m1-y\
&=dfrac-y1-y-dfrac-y^m+11-y\
endarray
$
so
$beginarray\
s_m(y)+dfracy1-y
&=dfracy^m+11-y\
endarray
$
Therefore
$sum_n=1^m frac(-1)^n-1x^n+dfrac-1/x1+1/x
=dfrac1(-x)^m+1(1+1/x)
$
or
$sum_n=1^m frac(-1)^n-1x^n-dfrac1x+1
=dfrac(-1)^m+1x^m(x+1)
$.
What is needed now is
to show that
$lim_m to infty dfrac1x^m(x+1)
=0$.
(This is from
"What is Mathematics")
Since $x > 1$,
$x = 1+z$
where $z > 0$.
By Bernoulli's inequality,
$x^m
=(1+z)^m
ge 1+mz
gt mz
=m(x-1)$,
so
$ dfrac1x^m(x+1)
lt dfrac1m(x-1)(x+1)
=dfrac1m(x^2-1)
$,
so to make
$dfrac1x^m(x+1)
lt epsilon
$
it is enough to take
$m
gt dfrac1epsilon(x^2-1)
$.
This is certainly
not the best $m$,
but it is
completely elementary.
answered Jul 16 at 20:19
marty cohen
69.3k446122
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's look at the
partial sums,
and let $y = -1/x$
so
$-1 < y < 0$..
$beginarray\
s_m(y)
&=sum_n=1^m (-1)^n-1(-y)^n\
&=sum_n=1^m (-1)^n-1(-1)^ny^n\
&=-sum_n=1^m y^n\
&=-ysum_n=0^m-1 y^n\
&=-ydfrac1-y^m1-y\
&=dfrac-y1-y-dfrac-y^m+11-y\
endarray
$
so
$beginarray\
s_m(y)+dfracy1-y
&=dfracy^m+11-y\
endarray
$
Therefore
$sum_n=1^m frac(-1)^n-1x^n+dfrac-1/x1+1/x
=dfrac1(-x)^m+1(1+1/x)
$
or
$sum_n=1^m frac(-1)^n-1x^n-dfrac1x+1
=dfrac(-1)^m+1x^m(x+1)
$.
What is needed now is
to show that
$lim_m to infty dfrac1x^m(x+1)
=0$.
(This is from
"What is Mathematics")
Since $x > 1$,
$x = 1+z$
where $z > 0$.
By Bernoulli's inequality,
$x^m
=(1+z)^m
ge 1+mz
gt mz
=m(x-1)$,
so
$ dfrac1x^m(x+1)
lt dfrac1m(x-1)(x+1)
=dfrac1m(x^2-1)
$,
so to make
$dfrac1x^m(x+1)
lt epsilon
$
it is enough to take
$m
gt dfrac1epsilon(x^2-1)
$.
This is certainly
not the best $m$,
but it is
completely elementary.
answered Jul 16 at 20:19
marty cohen
69.3k446122
Let's look at the
partial sums,
and let $y = -1/x$
so
$-1 < y < 0$..
$beginarray\
s_m(y)
&=sum_n=1^m (-1)^n-1(-y)^n\
&=sum_n=1^m (-1)^n-1(-1)^ny^n\
&=-sum_n=1^m y^n\
&=-ysum_n=0^m-1 y^n\
&=-ydfrac1-y^m1-y\
&=dfrac-y1-y-dfrac-y^m+11-y\
endarray
$
so
$beginarray\
s_m(y)+dfracy1-y
&=dfracy^m+11-y\
endarray
$
Therefore
$sum_n=1^m frac(-1)^n-1x^n+dfrac-1/x1+1/x
=dfrac1(-x)^m+1(1+1/x)
$
or
$sum_n=1^m frac(-1)^n-1x^n-dfrac1x+1
=dfrac(-1)^m+1x^m(x+1)
$.
What is needed now is
to show that
$lim_m to infty dfrac1x^m(x+1)
=0$.
(This is from
"What is Mathematics")
Since $x > 1$,
$x = 1+z$
where $z > 0$.
By Bernoulli's inequality,
$x^m
=(1+z)^m
ge 1+mz
gt mz
=m(x-1)$,
so
$ dfrac1x^m(x+1)
lt dfrac1m(x-1)(x+1)
=dfrac1m(x^2-1)
$,
so to make
$dfrac1x^m(x+1)
lt epsilon
$
it is enough to take
$m
gt dfrac1epsilon(x^2-1)
$.
This is certainly
not the best $m$,
but it is
completely elementary.
answered Jul 16 at 20:19
marty cohen
69.3k446122
answered Jul 16 at 20:19
marty cohen
69.3k446122
answered Jul 16 at 20:19
marty cohen
69.3k446122
answered Jul 16 at 20:19
marty cohen
69.3k446122
69.3k446122
add a comment |Â
add a comment |Â
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What convergence tests do you know? Since this summand is a simple ratio, there is an "obvious" choice for a test to try.
– Eric Towers
Jul 16 at 17:59