If $lambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$, for $n in mathbb{N}$, then $,lim_{n to infty}...
up vote
1
down vote
favorite
If $displaystylelambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$ for $n in mathbb{N}$. Then prove that $lim_{n to infty} (lambda_{n})^{1/n}=1.$
$$lambda_n=int_{0}^{1} frac{dt}{(1+t)^n}= frac{2^{1-n}}{1-n}-frac{1}{1-n}$$
Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.
calculus real-analysis limits definite-integrals limits-without-lhopital
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 10:34
ramanujan
670713
add a comment |
up vote
1
down vote
favorite
If $displaystylelambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$ for $n in mathbb{N}$. Then prove that $lim_{n to infty} (lambda_{n})^{1/n}=1.$
$$lambda_n=int_{0}^{1} frac{dt}{(1+t)^n}= frac{2^{1-n}}{1-n}-frac{1}{1-n}$$
Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.
calculus real-analysis limits definite-integrals limits-without-lhopital
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 10:34
ramanujan
670713
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $displaystylelambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$ for $n in mathbb{N}$. Then prove that $lim_{n to infty} (lambda_{n})^{1/n}=1.$
$$lambda_n=int_{0}^{1} frac{dt}{(1+t)^n}= frac{2^{1-n}}{1-n}-frac{1}{1-n}$$
Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.
calculus real-analysis limits definite-integrals limits-without-lhopital
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 10:34
ramanujan
670713
If $displaystylelambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$ for $n in mathbb{N}$. Then prove that $lim_{n to infty} (lambda_{n})^{1/n}=1.$
$$lambda_n=int_{0}^{1} frac{dt}{(1+t)^n}= frac{2^{1-n}}{1-n}-frac{1}{1-n}$$
Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.
calculus real-analysis limits definite-integrals limits-without-lhopital
calculus real-analysis limits definite-integrals limits-without-lhopital
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 10:34
ramanujan
670713
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62.2k1383162
asked Nov 19 at 10:34
ramanujan
670713
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62.2k1383162
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62.2k1383162
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62.2k1383162
62.2k1383162
asked Nov 19 at 10:34
ramanujan
670713
asked Nov 19 at 10:34
ramanujan
670713
asked Nov 19 at 10:34
ramanujan
670713
670713
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Actually,
$$
int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
$$
and hence, for all $n>1$
$$
frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
$$
Next, observe that
$$
lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
$$
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62.2k1383162
thank you. I wonder how such tricks come in your mind.
– ramanujan
Nov 19 at 11:05
1
@ramanujan: the old sage said Calculus is all about elementary inequalities.
– Jack D'Aurizio
Nov 19 at 13:31
add a comment |
up vote
2
down vote
In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,
$$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
by the inequality between means.
answered Nov 19 at 13:45
Jack D'Aurizio
284k33275654
Thanks for general result
– ramanujan
Nov 19 at 14:18
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Actually,
$$
int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
$$
and hence, for all $n>1$
$$
frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
$$
Next, observe that
$$
lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
$$
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62.2k1383162
thank you. I wonder how such tricks come in your mind.
– ramanujan
Nov 19 at 11:05
1
@ramanujan: the old sage said Calculus is all about elementary inequalities.
– Jack D'Aurizio
Nov 19 at 13:31
add a comment |
up vote
5
down vote
accepted
Actually,
$$
int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
$$
and hence, for all $n>1$
$$
frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
$$
Next, observe that
$$
lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
$$
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62.2k1383162
thank you. I wonder how such tricks come in your mind.
– ramanujan
Nov 19 at 11:05
1
@ramanujan: the old sage said Calculus is all about elementary inequalities.
– Jack D'Aurizio
Nov 19 at 13:31
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Actually,
$$
int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
$$
and hence, for all $n>1$
$$
frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
$$
Next, observe that
$$
lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
$$
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62.2k1383162
Actually,
$$
int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
$$
and hence, for all $n>1$
$$
frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
$$
Next, observe that
$$
lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
$$
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62.2k1383162
62.2k1383162
thank you. I wonder how such tricks come in your mind.
– ramanujan
Nov 19 at 11:05
1
@ramanujan: the old sage said Calculus is all about elementary inequalities.
– Jack D'Aurizio
Nov 19 at 13:31
add a comment |
thank you. I wonder how such tricks come in your mind.
– ramanujan
Nov 19 at 11:05
1
@ramanujan: the old sage said Calculus is all about elementary inequalities.
– Jack D'Aurizio
Nov 19 at 13:31
thank you. I wonder how such tricks come in your mind.
– ramanujan
Nov 19 at 11:05
thank you. I wonder how such tricks come in your mind.
– ramanujan
Nov 19 at 11:05
1
1
@ramanujan: the old sage said Calculus is all about elementary inequalities.
– Jack D'Aurizio
Nov 19 at 13:31
@ramanujan: the old sage said Calculus is all about elementary inequalities.
– Jack D'Aurizio
Nov 19 at 13:31
add a comment |
up vote
2
down vote
In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,
$$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
by the inequality between means.
answered Nov 19 at 13:45
Jack D'Aurizio
284k33275654
Thanks for general result
– ramanujan
Nov 19 at 14:18
add a comment |
up vote
2
down vote
In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,
$$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
by the inequality between means.
answered Nov 19 at 13:45
Jack D'Aurizio
284k33275654
Thanks for general result
– ramanujan
Nov 19 at 14:18
add a comment |
up vote
2
down vote
up vote
2
down vote
In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,
$$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
by the inequality between means.
answered Nov 19 at 13:45
Jack D'Aurizio
284k33275654
In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,
$$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
by the inequality between means.
answered Nov 19 at 13:45
Jack D'Aurizio
284k33275654
answered Nov 19 at 13:45
Jack D'Aurizio
284k33275654
answered Nov 19 at 13:45
Jack D'Aurizio
284k33275654
answered Nov 19 at 13:45
Jack D'Aurizio
284k33275654
284k33275654
Thanks for general result
– ramanujan
Nov 19 at 14:18
add a comment |
Thanks for general result
– ramanujan
Nov 19 at 14:18
Thanks for general result
– ramanujan
Nov 19 at 14:18
Thanks for general result
– ramanujan
Nov 19 at 14:18
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004767%2fif-lambda-n-int-01-fracdt1tn-for-n-in-mathbbn-then%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
