If $lambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$, for $n in mathbb{N}$, then $,lim_{n to infty}...
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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
62k1383161
asked Nov 19 at 10:34
ramanujan
668713
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
62k1383161
asked Nov 19 at 10:34
ramanujan
668713
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
62k1383161
asked Nov 19 at 10:34
ramanujan
668713
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
62k1383161
asked Nov 19 at 10:34
ramanujan
668713
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62k1383161
asked Nov 19 at 10:34
ramanujan
668713
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62k1383161
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62k1383161
edited Nov 19 at 12:01
Yiorgos S. Smyrlis
62k1383161
62k1383161
asked Nov 19 at 10:34
ramanujan
668713
asked Nov 19 at 10:34
ramanujan
668713
asked Nov 19 at 10:34
ramanujan
668713
668713
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
62k1383161
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
62k1383161
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
62k1383161
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
62k1383161
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
62k1383161
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62k1383161
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62k1383161
answered Nov 19 at 10:40
Yiorgos S. Smyrlis
62k1383161
62k1383161
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 |
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