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limit of the sequence $(n+1)^alpha - n^alpha$
The Next CEO of Stack OverflowFor $alpha in (0,frac32)$, define $x_n=(n+1)^alpha-n^alpha$. Then $lim_n to inftyx_n$ is?Doubly infinite sequence limitEvaluate the limit using only the following resultsDoes this sequence converge? If yes, what is the limit?Limit of the fraction $dfracn(n+1)^alphasum_k=1^nk^alpha$A limit like a Riemann sumFor $alpha in (0,frac32)$, define $x_n=(n+1)^alpha-n^alpha$. Then $lim_n to inftyx_n$ is?What is $lim_alpharightarrow0 left(alphalog^2alpharight)$?Does the series admit a limit as $alphato 0$?Convergence of a series of a sequence of powersProve a sequence is a $O(1/n^3)$
$begingroup$
Let $u_n$ be defined as : $u_n = (n+1)^alpha - n^alpha$.
Now I would like to compute $lim_n to infty u_n$ (this limit depends on the choice of $alpha$).
Here is what I found so far :
If $alpha in [0,1[$ then $lim_n to infty u_n = 0$. If $alpha = 1$ then $lim_n to infty u_n = 1$. If $alpha geq 2$, $lim_n to infty u_n = +infty$.
Now for $alpha = 1$ and $alpha geq 2$ it's really easy to get the result. For $alpha in [0,1)$ I found it hard and here is what I do :
we use the mean value theorem to say that :
$$mid (n+1)^alpha - n^alpha mid leq alpha n^alpha-1 to 0$$
And we get the desired result. Now my question is :
1- Is it possible to get the limit when $alphain [0,1)$ with an other technique (without using the mean value theorem) ?
2- How to get the limit when $alpha in (1, 2)$ ?
Thank you !
real-analysis sequences-and-series limits
asked Mar 18 at 12:23
mouargmouargmouargmouarg
1145
$endgroup$
|
show 4 more comments
$begingroup$
Let $u_n$ be defined as : $u_n = (n+1)^alpha - n^alpha$.
Now I would like to compute $lim_n to infty u_n$ (this limit depends on the choice of $alpha$).
Here is what I found so far :
If $alpha in [0,1[$ then $lim_n to infty u_n = 0$. If $alpha = 1$ then $lim_n to infty u_n = 1$. If $alpha geq 2$, $lim_n to infty u_n = +infty$.
Now for $alpha = 1$ and $alpha geq 2$ it's really easy to get the result. For $alpha in [0,1)$ I found it hard and here is what I do :
we use the mean value theorem to say that :
$$mid (n+1)^alpha - n^alpha mid leq alpha n^alpha-1 to 0$$
And we get the desired result. Now my question is :
1- Is it possible to get the limit when $alphain [0,1)$ with an other technique (without using the mean value theorem) ?
2- How to get the limit when $alpha in (1, 2)$ ?
Thank you !
real-analysis sequences-and-series limits
asked Mar 18 at 12:23
mouargmouargmouargmouarg
1145
$endgroup$
1
$begingroup$
$f(x) =x^a$ then $f'=ax^a-1$
$endgroup$
– dmtri
Mar 18 at 12:34
2
$begingroup$
The $alpha n^1-alpha$ above should be $alpha n^alpha-1$
$endgroup$
– robjohn♦
Mar 18 at 12:34
2
$begingroup$
May mean $[0,1)$
$endgroup$
– J. W. Tanner
Mar 18 at 12:42
1
$begingroup$
@robjohn Thank you for noticing this. It shoud be correct now
$endgroup$
– mouargmouarg
Mar 18 at 12:48
1
$begingroup$
@dmtri Thank you for catching the mistake. And yes it mean $[0,1)$.
$endgroup$
– mouargmouarg
Mar 18 at 12:49
|
show 4 more comments
$begingroup$
Let $u_n$ be defined as : $u_n = (n+1)^alpha - n^alpha$.
Now I would like to compute $lim_n to infty u_n$ (this limit depends on the choice of $alpha$).
Here is what I found so far :
If $alpha in [0,1[$ then $lim_n to infty u_n = 0$. If $alpha = 1$ then $lim_n to infty u_n = 1$. If $alpha geq 2$, $lim_n to infty u_n = +infty$.
Now for $alpha = 1$ and $alpha geq 2$ it's really easy to get the result. For $alpha in [0,1)$ I found it hard and here is what I do :
we use the mean value theorem to say that :
$$mid (n+1)^alpha - n^alpha mid leq alpha n^alpha-1 to 0$$
And we get the desired result. Now my question is :
1- Is it possible to get the limit when $alphain [0,1)$ with an other technique (without using the mean value theorem) ?
2- How to get the limit when $alpha in (1, 2)$ ?
Thank you !
real-analysis sequences-and-series limits
asked Mar 18 at 12:23
mouargmouargmouargmouarg
1145
$endgroup$
Let $u_n$ be defined as : $u_n = (n+1)^alpha - n^alpha$.
Now I would like to compute $lim_n to infty u_n$ (this limit depends on the choice of $alpha$).
Here is what I found so far :
If $alpha in [0,1[$ then $lim_n to infty u_n = 0$. If $alpha = 1$ then $lim_n to infty u_n = 1$. If $alpha geq 2$, $lim_n to infty u_n = +infty$.
Now for $alpha = 1$ and $alpha geq 2$ it's really easy to get the result. For $alpha in [0,1)$ I found it hard and here is what I do :
we use the mean value theorem to say that :
$$mid (n+1)^alpha - n^alpha mid leq alpha n^alpha-1 to 0$$
And we get the desired result. Now my question is :
1- Is it possible to get the limit when $alphain [0,1)$ with an other technique (without using the mean value theorem) ?
2- How to get the limit when $alpha in (1, 2)$ ?
Thank you !
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
asked Mar 18 at 12:23
mouargmouargmouargmouarg
1145
asked Mar 18 at 12:23
mouargmouargmouargmouarg
1145
edited Mar 18 at 12:48
mouargmouarg
asked Mar 18 at 12:23
mouargmouargmouargmouarg
1145
asked Mar 18 at 12:23
mouargmouargmouargmouarg
1145
asked Mar 18 at 12:23
mouargmouargmouargmouarg
1145
1145
1
$begingroup$
$f(x) =x^a$ then $f'=ax^a-1$
$endgroup$
– dmtri
Mar 18 at 12:34
2
$begingroup$
The $alpha n^1-alpha$ above should be $alpha n^alpha-1$
$endgroup$
– robjohn♦
Mar 18 at 12:34
2
$begingroup$
May mean $[0,1)$
$endgroup$
– J. W. Tanner
Mar 18 at 12:42
1
$begingroup$
@robjohn Thank you for noticing this. It shoud be correct now
$endgroup$
– mouargmouarg
Mar 18 at 12:48
1
$begingroup$
@dmtri Thank you for catching the mistake. And yes it mean $[0,1)$.
$endgroup$
– mouargmouarg
Mar 18 at 12:49
|
show 4 more comments
1
$begingroup$
$f(x) =x^a$ then $f'=ax^a-1$
$endgroup$
– dmtri
Mar 18 at 12:34
2
$begingroup$
The $alpha n^1-alpha$ above should be $alpha n^alpha-1$
$endgroup$
– robjohn♦
Mar 18 at 12:34
2
$begingroup$
May mean $[0,1)$
$endgroup$
– J. W. Tanner
Mar 18 at 12:42
1
$begingroup$
@robjohn Thank you for noticing this. It shoud be correct now
$endgroup$
– mouargmouarg
Mar 18 at 12:48
1
$begingroup$
@dmtri Thank you for catching the mistake. And yes it mean $[0,1)$.
$endgroup$
– mouargmouarg
Mar 18 at 12:49
1
1
$begingroup$
$f(x) =x^a$ then $f'=ax^a-1$
$endgroup$
– dmtri
Mar 18 at 12:34
$begingroup$
$f(x) =x^a$ then $f'=ax^a-1$
$endgroup$
– dmtri
Mar 18 at 12:34
2
2
$begingroup$
The $alpha n^1-alpha$ above should be $alpha n^alpha-1$
$endgroup$
– robjohn♦
Mar 18 at 12:34
$begingroup$
The $alpha n^1-alpha$ above should be $alpha n^alpha-1$
$endgroup$
– robjohn♦
Mar 18 at 12:34
2
2
$begingroup$
May mean $[0,1)$
$endgroup$
– J. W. Tanner
Mar 18 at 12:42
$begingroup$
May mean $[0,1)$
$endgroup$
– J. W. Tanner
Mar 18 at 12:42
1
1
$begingroup$
@robjohn Thank you for noticing this. It shoud be correct now
$endgroup$
– mouargmouarg
Mar 18 at 12:48
$begingroup$
@robjohn Thank you for noticing this. It shoud be correct now
$endgroup$
– mouargmouarg
Mar 18 at 12:48
1
1
$begingroup$
@dmtri Thank you for catching the mistake. And yes it mean $[0,1)$.
$endgroup$
– mouargmouarg
Mar 18 at 12:49
$begingroup$
@dmtri Thank you for catching the mistake. And yes it mean $[0,1)$.
$endgroup$
– mouargmouarg
Mar 18 at 12:49
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Notice that
$$alpha n^alpha-1 leq u_nleq alpha(n+1)^alpha-1.$$
Hence by the Sandwich theorem,
$$lim alpha n^alpha-1leqlim u_nleqlimalpha n^alpha-1$$
since $lim alpha n^alpha-1=limalpha (n+1)^alpha-1$. It remains to find $limalpha n^alpha-1$.
answered Mar 18 at 13:02
YiFanYiFan
4,8511727
$endgroup$
$begingroup$
How did you obtain that estimate? Note that OP asked for a technique “without using the mean value theorem.”
$endgroup$
– Martin R
Mar 18 at 13:39
$begingroup$
@MartinR There's no need for MVT. Just note that the derivative of the function is monotonous (by looking at the second derivative) so that we have the desired bound by the Intermediate Value Theorem.
$endgroup$
– YiFan
Mar 18 at 13:43
add a comment |
$begingroup$
If $alphale 0$ the limit is $0$.
Suppose $alpha > 0$.
We write: $$u_n=n^alpha(1-frac1n)^alpha=n^alphaleft(-fracalphan+o(frac1n^2)right)$$
$$=alpha n^alpha-1+o(n^alpha-2)$$
We conclude that:
If $0<alpha<1, u_nlongrightarrow 0$
If $alpha>1, u_nlongrightarrow infty$
if $alpha=1$ the limit is 1.
answered Mar 18 at 14:09
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Notice that
$$alpha n^alpha-1 leq u_nleq alpha(n+1)^alpha-1.$$
Hence by the Sandwich theorem,
$$lim alpha n^alpha-1leqlim u_nleqlimalpha n^alpha-1$$
since $lim alpha n^alpha-1=limalpha (n+1)^alpha-1$. It remains to find $limalpha n^alpha-1$.
answered Mar 18 at 13:02
YiFanYiFan
4,8511727
$endgroup$
$begingroup$
How did you obtain that estimate? Note that OP asked for a technique “without using the mean value theorem.”
$endgroup$
– Martin R
Mar 18 at 13:39
$begingroup$
@MartinR There's no need for MVT. Just note that the derivative of the function is monotonous (by looking at the second derivative) so that we have the desired bound by the Intermediate Value Theorem.
$endgroup$
– YiFan
Mar 18 at 13:43
add a comment |
$begingroup$
Hint: Notice that
$$alpha n^alpha-1 leq u_nleq alpha(n+1)^alpha-1.$$
Hence by the Sandwich theorem,
$$lim alpha n^alpha-1leqlim u_nleqlimalpha n^alpha-1$$
since $lim alpha n^alpha-1=limalpha (n+1)^alpha-1$. It remains to find $limalpha n^alpha-1$.
answered Mar 18 at 13:02
YiFanYiFan
4,8511727
$endgroup$
$begingroup$
How did you obtain that estimate? Note that OP asked for a technique “without using the mean value theorem.”
$endgroup$
– Martin R
Mar 18 at 13:39
$begingroup$
@MartinR There's no need for MVT. Just note that the derivative of the function is monotonous (by looking at the second derivative) so that we have the desired bound by the Intermediate Value Theorem.
$endgroup$
– YiFan
Mar 18 at 13:43
add a comment |
$begingroup$
Hint: Notice that
$$alpha n^alpha-1 leq u_nleq alpha(n+1)^alpha-1.$$
Hence by the Sandwich theorem,
$$lim alpha n^alpha-1leqlim u_nleqlimalpha n^alpha-1$$
since $lim alpha n^alpha-1=limalpha (n+1)^alpha-1$. It remains to find $limalpha n^alpha-1$.
answered Mar 18 at 13:02
YiFanYiFan
4,8511727
$endgroup$
Hint: Notice that
$$alpha n^alpha-1 leq u_nleq alpha(n+1)^alpha-1.$$
Hence by the Sandwich theorem,
$$lim alpha n^alpha-1leqlim u_nleqlimalpha n^alpha-1$$
since $lim alpha n^alpha-1=limalpha (n+1)^alpha-1$. It remains to find $limalpha n^alpha-1$.
answered Mar 18 at 13:02
YiFanYiFan
4,8511727
answered Mar 18 at 13:02
YiFanYiFan
4,8511727
answered Mar 18 at 13:02
YiFanYiFan
4,8511727
answered Mar 18 at 13:02
YiFanYiFan
4,8511727
4,8511727
$begingroup$
How did you obtain that estimate? Note that OP asked for a technique “without using the mean value theorem.”
$endgroup$
– Martin R
Mar 18 at 13:39
$begingroup$
@MartinR There's no need for MVT. Just note that the derivative of the function is monotonous (by looking at the second derivative) so that we have the desired bound by the Intermediate Value Theorem.
$endgroup$
– YiFan
Mar 18 at 13:43
add a comment |
$begingroup$
How did you obtain that estimate? Note that OP asked for a technique “without using the mean value theorem.”
$endgroup$
– Martin R
Mar 18 at 13:39
$begingroup$
@MartinR There's no need for MVT. Just note that the derivative of the function is monotonous (by looking at the second derivative) so that we have the desired bound by the Intermediate Value Theorem.
$endgroup$
– YiFan
Mar 18 at 13:43
$begingroup$
How did you obtain that estimate? Note that OP asked for a technique “without using the mean value theorem.”
$endgroup$
– Martin R
Mar 18 at 13:39
$begingroup$
How did you obtain that estimate? Note that OP asked for a technique “without using the mean value theorem.”
$endgroup$
– Martin R
Mar 18 at 13:39
$begingroup$
@MartinR There's no need for MVT. Just note that the derivative of the function is monotonous (by looking at the second derivative) so that we have the desired bound by the Intermediate Value Theorem.
$endgroup$
– YiFan
Mar 18 at 13:43
$begingroup$
@MartinR There's no need for MVT. Just note that the derivative of the function is monotonous (by looking at the second derivative) so that we have the desired bound by the Intermediate Value Theorem.
$endgroup$
– YiFan
Mar 18 at 13:43
add a comment |
$begingroup$
If $alphale 0$ the limit is $0$.
Suppose $alpha > 0$.
We write: $$u_n=n^alpha(1-frac1n)^alpha=n^alphaleft(-fracalphan+o(frac1n^2)right)$$
$$=alpha n^alpha-1+o(n^alpha-2)$$
We conclude that:
If $0<alpha<1, u_nlongrightarrow 0$
If $alpha>1, u_nlongrightarrow infty$
if $alpha=1$ the limit is 1.
answered Mar 18 at 14:09
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
$endgroup$
add a comment |
$begingroup$
If $alphale 0$ the limit is $0$.
Suppose $alpha > 0$.
We write: $$u_n=n^alpha(1-frac1n)^alpha=n^alphaleft(-fracalphan+o(frac1n^2)right)$$
$$=alpha n^alpha-1+o(n^alpha-2)$$
We conclude that:
If $0<alpha<1, u_nlongrightarrow 0$
If $alpha>1, u_nlongrightarrow infty$
if $alpha=1$ the limit is 1.
answered Mar 18 at 14:09
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
$endgroup$
add a comment |
$begingroup$
If $alphale 0$ the limit is $0$.
Suppose $alpha > 0$.
We write: $$u_n=n^alpha(1-frac1n)^alpha=n^alphaleft(-fracalphan+o(frac1n^2)right)$$
$$=alpha n^alpha-1+o(n^alpha-2)$$
We conclude that:
If $0<alpha<1, u_nlongrightarrow 0$
If $alpha>1, u_nlongrightarrow infty$
if $alpha=1$ the limit is 1.
answered Mar 18 at 14:09
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
$endgroup$
If $alphale 0$ the limit is $0$.
Suppose $alpha > 0$.
We write: $$u_n=n^alpha(1-frac1n)^alpha=n^alphaleft(-fracalphan+o(frac1n^2)right)$$
$$=alpha n^alpha-1+o(n^alpha-2)$$
We conclude that:
If $0<alpha<1, u_nlongrightarrow 0$
If $alpha>1, u_nlongrightarrow infty$
if $alpha=1$ the limit is 1.
answered Mar 18 at 14:09
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
edited Mar 18 at 14:19
answered Mar 18 at 14:09
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
answered Mar 18 at 14:09
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
answered Mar 18 at 14:09
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
1
add a comment |
add a comment |
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1
$begingroup$
$f(x) =x^a$ then $f'=ax^a-1$
$endgroup$
– dmtri
Mar 18 at 12:34
2
$begingroup$
The $alpha n^1-alpha$ above should be $alpha n^alpha-1$
$endgroup$
– robjohn♦
Mar 18 at 12:34
2
$begingroup$
May mean $[0,1)$
$endgroup$
– J. W. Tanner
Mar 18 at 12:42
1
$begingroup$
@robjohn Thank you for noticing this. It shoud be correct now
$endgroup$
– mouargmouarg
Mar 18 at 12:48
1
$begingroup$
@dmtri Thank you for catching the mistake. And yes it mean $[0,1)$.
$endgroup$
– mouargmouarg
Mar 18 at 12:49