Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$ ?
My attempt: I was thinking about $M_n$ test
$$M_n = sup_x in [1,a] | n( sqrt [n]x-1)- 0|=sup_x in [1,a] |n(e^frac1nlog x -1)- 0|$$
After that i can not able to proceed further. Thanks.
real-analysis
edited Sep 4 at 11:25
Robert Z
85.7k1055124
asked Sep 4 at 10:41
stupid
712112
add a comment |Â
up vote
0
down vote
favorite
Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$ ?
My attempt: I was thinking about $M_n$ test
$$M_n = sup_x in [1,a] | n( sqrt [n]x-1)- 0|=sup_x in [1,a] |n(e^frac1nlog x -1)- 0|$$
After that i can not able to proceed further. Thanks.
real-analysis
edited Sep 4 at 11:25
Robert Z
85.7k1055124
asked Sep 4 at 10:41
stupid
712112
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$ ?
My attempt: I was thinking about $M_n$ test
$$M_n = sup_x in [1,a] | n( sqrt [n]x-1)- 0|=sup_x in [1,a] |n(e^frac1nlog x -1)- 0|$$
After that i can not able to proceed further. Thanks.
real-analysis
edited Sep 4 at 11:25
Robert Z
85.7k1055124
asked Sep 4 at 10:41
stupid
712112
Is the sequence $f_n(x) = n(sqrt [n]x-1)$ converges uniformly on $[1,a]$ where $a >1$ ?
My attempt: I was thinking about $M_n$ test
$$M_n = sup_x in [1,a] | n( sqrt [n]x-1)- 0|=sup_x in [1,a] |n(e^frac1nlog x -1)- 0|$$
After that i can not able to proceed further. Thanks.
real-analysis
real-analysis
edited Sep 4 at 11:25
Robert Z
85.7k1055124
asked Sep 4 at 10:41
stupid
712112
edited Sep 4 at 11:25
Robert Z
85.7k1055124
asked Sep 4 at 10:41
stupid
712112
edited Sep 4 at 11:25
Robert Z
85.7k1055124
edited Sep 4 at 11:25
Robert Z
85.7k1055124
edited Sep 4 at 11:25
Robert Z
85.7k1055124
85.7k1055124
asked Sep 4 at 10:41
stupid
712112
asked Sep 4 at 10:41
stupid
712112
asked Sep 4 at 10:41
stupid
712112
712112
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
$$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
Hence, following your approach, you should consider
$$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
Can you take it from here?
Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.
answered Sep 4 at 11:03
Robert Z
85.7k1055124
thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
– stupid
Sep 4 at 11:13
@stupid See my edit.
– Robert Z
Sep 4 at 11:45
thanksu @Robert
– stupid
Sep 4 at 12:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
$$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
Hence, following your approach, you should consider
$$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
Can you take it from here?
Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.
answered Sep 4 at 11:03
Robert Z
85.7k1055124
thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
– stupid
Sep 4 at 11:13
@stupid See my edit.
– Robert Z
Sep 4 at 11:45
thanksu @Robert
– stupid
Sep 4 at 12:03
add a comment |Â
up vote
3
down vote
accepted
Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
$$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
Hence, following your approach, you should consider
$$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
Can you take it from here?
Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.
answered Sep 4 at 11:03
Robert Z
85.7k1055124
thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
– stupid
Sep 4 at 11:13
@stupid See my edit.
– Robert Z
Sep 4 at 11:45
thanksu @Robert
– stupid
Sep 4 at 12:03
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
$$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
Hence, following your approach, you should consider
$$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
Can you take it from here?
Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.
answered Sep 4 at 11:03
Robert Z
85.7k1055124
Hint. The pointwise limit in $[1,a]$ is $ln(x)$ (it is not $0$):
$$lim_ntoinftyf_n(x)=lim_ntoinfty frac e^ln(x)/n-11/n=ln(x).$$
Hence, following your approach, you should consider
$$sup_x in [1,a] |n(e^ln(x)/n -1)- ln(x)|=
nsup_t in [0,fracln(a)n] |e^t -1- t|.$$
Can you take it from here?
Alternatively, you may that for any $xin [1,a]$ the sequence $f_n(x)_ngeq 1$ is decreasing. Then use the Dini's Theorem.
answered Sep 4 at 11:03
Robert Z
85.7k1055124
edited Sep 4 at 11:45
answered Sep 4 at 11:03
Robert Z
85.7k1055124
answered Sep 4 at 11:03
Robert Z
85.7k1055124
answered Sep 4 at 11:03
Robert Z
85.7k1055124
85.7k1055124
thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
– stupid
Sep 4 at 11:13
@stupid See my edit.
– Robert Z
Sep 4 at 11:45
thanksu @Robert
– stupid
Sep 4 at 12:03
add a comment |Â
thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
– stupid
Sep 4 at 11:13
@stupid See my edit.
– Robert Z
Sep 4 at 11:45
thanksu @Robert
– stupid
Sep 4 at 12:03
thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
– stupid
Sep 4 at 11:13
thanks @Robert oh yaa i misses that...but $sup_x in [1,a] |e^frac1nlog x -1)- logx|$.... how can i proceed further??
– stupid
Sep 4 at 11:13
@stupid See my edit.
– Robert Z
Sep 4 at 11:45
@stupid See my edit.
– Robert Z
Sep 4 at 11:45
thanksu @Robert
– stupid
Sep 4 at 12:03
thanksu @Robert
– stupid
Sep 4 at 12:03
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904885%2fis-the-sequence-f-nx-n-sqrt-nx-1-converges-uniformly-on-1-a-wher%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
