Find limit $lim_nto inftyfracsum_k=1^n k^nn^n$.

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up vote
5
down vote

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Now I want to find the limit
$$lim_ntoinftyfracsumlimits_k=1^n k^nn^n.$$
I try to use the Stolz theorem as follows:
$$lim_nto inftyfracsum_k=1^n k^nn^n=
lim_nto inftyfracsum_k=1^n+1 k^n+1-sum_k=1^n k^n(n+1)^n+1-n^n$$
$$=lim_ntoinftyfrac(n+1)^n+1+sum_k=1^n(k^n+1-k^n)(n+1)^n+1-n^n$$
$$=1+lim_ntoinftyfracsum_k=1^n(k^n+1-k^n)(n+1)^n+1.$$
It seems to deal with the summation like this form:
$$sum_k=1^nk^p,textwith p=n,n+1.$$
I have no way to deal this summation, any help and hint will welcome!

sequences-and-series number-theory limits

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asked Jul 31 at 14:05

Riemann

1,9891217

add a comment | 

up vote
5
down vote

favorite

Now I want to find the limit
$$lim_ntoinftyfracsumlimits_k=1^n k^nn^n.$$
I try to use the Stolz theorem as follows:
$$lim_nto inftyfracsum_k=1^n k^nn^n=
lim_nto inftyfracsum_k=1^n+1 k^n+1-sum_k=1^n k^n(n+1)^n+1-n^n$$
$$=lim_ntoinftyfrac(n+1)^n+1+sum_k=1^n(k^n+1-k^n)(n+1)^n+1-n^n$$
$$=1+lim_ntoinftyfracsum_k=1^n(k^n+1-k^n)(n+1)^n+1.$$
It seems to deal with the summation like this form:
$$sum_k=1^nk^p,textwith p=n,n+1.$$
I have no way to deal this summation, any help and hint will welcome!

sequences-and-series number-theory limits

share|cite|improve this question

asked Jul 31 at 14:05

Riemann

1,9891217

add a comment | 

up vote
5
down vote

favorite

up vote
5
down vote

favorite

Now I want to find the limit
$$lim_ntoinftyfracsumlimits_k=1^n k^nn^n.$$
I try to use the Stolz theorem as follows:
$$lim_nto inftyfracsum_k=1^n k^nn^n=
lim_nto inftyfracsum_k=1^n+1 k^n+1-sum_k=1^n k^n(n+1)^n+1-n^n$$
$$=lim_ntoinftyfrac(n+1)^n+1+sum_k=1^n(k^n+1-k^n)(n+1)^n+1-n^n$$
$$=1+lim_ntoinftyfracsum_k=1^n(k^n+1-k^n)(n+1)^n+1.$$
It seems to deal with the summation like this form:
$$sum_k=1^nk^p,textwith p=n,n+1.$$
I have no way to deal this summation, any help and hint will welcome!

sequences-and-series number-theory limits

share|cite|improve this question

asked Jul 31 at 14:05

Riemann

1,9891217

Now I want to find the limit
$$lim_ntoinftyfracsumlimits_k=1^n k^nn^n.$$
I try to use the Stolz theorem as follows:
$$lim_nto inftyfracsum_k=1^n k^nn^n=
lim_nto inftyfracsum_k=1^n+1 k^n+1-sum_k=1^n k^n(n+1)^n+1-n^n$$
$$=lim_ntoinftyfrac(n+1)^n+1+sum_k=1^n(k^n+1-k^n)(n+1)^n+1-n^n$$
$$=1+lim_ntoinftyfracsum_k=1^n(k^n+1-k^n)(n+1)^n+1.$$
It seems to deal with the summation like this form:
$$sum_k=1^nk^p,textwith p=n,n+1.$$
I have no way to deal this summation, any help and hint will welcome!

sequences-and-series number-theory limits

share|cite|improve this question

asked Jul 31 at 14:05

Riemann

1,9891217

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 14:05

Riemann

1,9891217

asked Jul 31 at 14:05

Riemann

1,9891217

asked Jul 31 at 14:05

Riemann

1,9891217

1,9891217

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
7
down vote



accepted

Hint: For fixed $k,$ $$frac(n-k)^nn^n=left(1-fracknright)^nto e^-k$$

---

Details:

Note that for $k=0...,n-1$ $$logleft(1-fracknright)=-frackn-frack^22n^2-cdotsleq -frackn$$

So $$left(1-fracknright)^n leq e^-k$$

Then apply the dominated convergence theorem by defining:

$$f_n(k)=begincases0 & kgeq n\ left(1-fracknright)^n& 0leq k<nendcases$$

Then $|f_n(k)|leq g(k)=e^-k$ and for any $k,$ $lim_ktoinftyf_n(k)to g(k).$

share|cite|improve this answer

edited Jul 31 at 22:05

answered Jul 31 at 14:10

Thomas Andrews

128k10143284

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How to deal with the double limit??
  – Riemann
  Jul 31 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the value of limit is ?.Numeric says: $approx1.58$
  – Mariusz Iwaniuk
  Jul 31 at 14:34
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  the limit is $fracee-1approx1.58$.
  – Riemann
  Jul 31 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Could someone add a whole solution? I am interested in this limit aswell but I am not sure how to use the given hint to find the correct value.
  – mrtaurho
  Jul 31 at 14:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @mrtaurho 1+1/e+1/e^2+・・・+1/e^n+・・・$=frac11-1/e=fracee-1$
  – Takahiro Waki
  Jul 31 at 14:57
  

  
  
  
  

 | 
show 1 more comment

up vote
7
down vote

We have that

$$fracsumlimits_k=1^n k^nn^n=sum_k=1^n left( frac k n right)^n=sum_k=0^n-1 left( frac n-k n right)^n=sum_k=0^n-1 left( 1-frac k n right)^n$$

then, following the suggestion given by ComplexYetTrivial, let consider

$$a_n=sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n$$

which is strictly increasing and bounded indeed by AM-GM we have

• $sqrt[n+1]left(1-fracknright)^n cdot 1 leq fracn
  left(1 - fracknright) + 1n+1 = 1 - frackn+1 $

and

• $left( 1-frac k n right)^n=e^nlog left( 1-frac k n
  right)=e^n left( -frac k n-frac k^2 2n^2-frac k^3
  3n^3-ldots right)le e^-k$

therefore by the monotone convergence theorem $a_n$ has finite limit and since for $k$ fixed

• $left( 1-frac k n right)^nto e^-k$

we have

$$lim_nto inftyfracsumlimits_k=1^n k^nn^n=lim_nto infty sum_k=0^n-1 left( 1-frac k n right)^n=lim_nto infty sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n=sum_k=0^inftye^-k=fracee-1$$

share|cite|improve this answer

edited Aug 1 at 21:08

answered Jul 31 at 14:46

gimusi

64.1k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) without Hints at last. :P
  – Mariusz Iwaniuk
  Jul 31 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk Thanks! The hint was already given! :)
  – gimusi
  Jul 31 at 15:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In your second line, the $O(frac1n)$ depends on $k$. Therefore you can’t factorize $O(frac1n)$ in the sum without an additional argument.
  – mathcounterexamples.net
  Jul 31 at 15:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net Yes I think you are right, it's a step I need to revise. Thanks
  – gimusi
  Jul 31 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could write the sum as $sum_k=0^infty left(1-fracknright)^n operatorname1(k<n)$, observe that the sequence $left[left(1-fracknright)^n operatorname1(k<n)right]_ninmathbbN$ is increasing and use the monotone convergence theorem to interchange summation and limit.
  – ComplexYetTrivial
  Jul 31 at 16:16
  
  

  
  
  
  

 | 
show 4 more comments

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

Hint: For fixed $k,$ $$frac(n-k)^nn^n=left(1-fracknright)^nto e^-k$$

---

Details:

Note that for $k=0...,n-1$ $$logleft(1-fracknright)=-frackn-frack^22n^2-cdotsleq -frackn$$

So $$left(1-fracknright)^n leq e^-k$$

Then apply the dominated convergence theorem by defining:

$$f_n(k)=begincases0 & kgeq n\ left(1-fracknright)^n& 0leq k<nendcases$$

Then $|f_n(k)|leq g(k)=e^-k$ and for any $k,$ $lim_ktoinftyf_n(k)to g(k).$

share|cite|improve this answer

edited Jul 31 at 22:05

answered Jul 31 at 14:10

Thomas Andrews

128k10143284

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How to deal with the double limit??
  – Riemann
  Jul 31 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the value of limit is ?.Numeric says: $approx1.58$
  – Mariusz Iwaniuk
  Jul 31 at 14:34
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  the limit is $fracee-1approx1.58$.
  – Riemann
  Jul 31 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Could someone add a whole solution? I am interested in this limit aswell but I am not sure how to use the given hint to find the correct value.
  – mrtaurho
  Jul 31 at 14:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @mrtaurho 1+1/e+1/e^2+・・・+1/e^n+・・・$=frac11-1/e=fracee-1$
  – Takahiro Waki
  Jul 31 at 14:57
  

  
  
  
  

 | 
show 1 more comment

up vote
7
down vote



accepted

Hint: For fixed $k,$ $$frac(n-k)^nn^n=left(1-fracknright)^nto e^-k$$

---

Details:

Note that for $k=0...,n-1$ $$logleft(1-fracknright)=-frackn-frack^22n^2-cdotsleq -frackn$$

So $$left(1-fracknright)^n leq e^-k$$

Then apply the dominated convergence theorem by defining:

$$f_n(k)=begincases0 & kgeq n\ left(1-fracknright)^n& 0leq k<nendcases$$

Then $|f_n(k)|leq g(k)=e^-k$ and for any $k,$ $lim_ktoinftyf_n(k)to g(k).$

share|cite|improve this answer

edited Jul 31 at 22:05

answered Jul 31 at 14:10

Thomas Andrews

128k10143284

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How to deal with the double limit??
  – Riemann
  Jul 31 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the value of limit is ?.Numeric says: $approx1.58$
  – Mariusz Iwaniuk
  Jul 31 at 14:34
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  the limit is $fracee-1approx1.58$.
  – Riemann
  Jul 31 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Could someone add a whole solution? I am interested in this limit aswell but I am not sure how to use the given hint to find the correct value.
  – mrtaurho
  Jul 31 at 14:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @mrtaurho 1+1/e+1/e^2+・・・+1/e^n+・・・$=frac11-1/e=fracee-1$
  – Takahiro Waki
  Jul 31 at 14:57
  

  
  
  
  

 | 
show 1 more comment

up vote
7
down vote



accepted

up vote
7
down vote



accepted

Hint: For fixed $k,$ $$frac(n-k)^nn^n=left(1-fracknright)^nto e^-k$$

---

Details:

Note that for $k=0...,n-1$ $$logleft(1-fracknright)=-frackn-frack^22n^2-cdotsleq -frackn$$

So $$left(1-fracknright)^n leq e^-k$$

Then apply the dominated convergence theorem by defining:

$$f_n(k)=begincases0 & kgeq n\ left(1-fracknright)^n& 0leq k<nendcases$$

Then $|f_n(k)|leq g(k)=e^-k$ and for any $k,$ $lim_ktoinftyf_n(k)to g(k).$

share|cite|improve this answer

edited Jul 31 at 22:05

answered Jul 31 at 14:10

Thomas Andrews

128k10143284

Hint: For fixed $k,$ $$frac(n-k)^nn^n=left(1-fracknright)^nto e^-k$$

---

Details:

Note that for $k=0...,n-1$ $$logleft(1-fracknright)=-frackn-frack^22n^2-cdotsleq -frackn$$

So $$left(1-fracknright)^n leq e^-k$$

Then apply the dominated convergence theorem by defining:

$$f_n(k)=begincases0 & kgeq n\ left(1-fracknright)^n& 0leq k<nendcases$$

Then $|f_n(k)|leq g(k)=e^-k$ and for any $k,$ $lim_ktoinftyf_n(k)to g(k).$

share|cite|improve this answer

edited Jul 31 at 22:05

answered Jul 31 at 14:10

Thomas Andrews

128k10143284

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 22:05

edited Jul 31 at 22:05

edited Jul 31 at 22:05

answered Jul 31 at 14:10

Thomas Andrews

128k10143284

answered Jul 31 at 14:10

Thomas Andrews

128k10143284

answered Jul 31 at 14:10

Thomas Andrews

128k10143284

128k10143284

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How to deal with the double limit??
  – Riemann
  Jul 31 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the value of limit is ?.Numeric says: $approx1.58$
  – Mariusz Iwaniuk
  Jul 31 at 14:34
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  the limit is $fracee-1approx1.58$.
  – Riemann
  Jul 31 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Could someone add a whole solution? I am interested in this limit aswell but I am not sure how to use the given hint to find the correct value.
  – mrtaurho
  Jul 31 at 14:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @mrtaurho 1+1/e+1/e^2+・・・+1/e^n+・・・$=frac11-1/e=fracee-1$
  – Takahiro Waki
  Jul 31 at 14:57
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How to deal with the double limit??
  – Riemann
  Jul 31 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  and the value of limit is ?.Numeric says: $approx1.58$
  – Mariusz Iwaniuk
  Jul 31 at 14:34
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  the limit is $fracee-1approx1.58$.
  – Riemann
  Jul 31 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Could someone add a whole solution? I am interested in this limit aswell but I am not sure how to use the given hint to find the correct value.
  – mrtaurho
  Jul 31 at 14:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @mrtaurho 1+1/e+1/e^2+・・・+1/e^n+・・・$=frac11-1/e=fracee-1$
  – Takahiro Waki
  Jul 31 at 14:57
  

  
  
  
  

How to deal with the double limit??
– Riemann
Jul 31 at 14:33

How to deal with the double limit??
– Riemann
Jul 31 at 14:33

and the value of limit is ?.Numeric says: $approx1.58$
– Mariusz Iwaniuk
Jul 31 at 14:34

and the value of limit is ?.Numeric says: $approx1.58$
– Mariusz Iwaniuk
Jul 31 at 14:34

2

2

the limit is $fracee-1approx1.58$.
– Riemann
Jul 31 at 14:39

the limit is $fracee-1approx1.58$.
– Riemann
Jul 31 at 14:39

1

1

Could someone add a whole solution? I am interested in this limit aswell but I am not sure how to use the given hint to find the correct value.
– mrtaurho
Jul 31 at 14:44

Could someone add a whole solution? I am interested in this limit aswell but I am not sure how to use the given hint to find the correct value.
– mrtaurho
Jul 31 at 14:44

2

2

@mrtaurho 1+1/e+1/e^2+・・・+1/e^n+・・・$=frac11-1/e=fracee-1$
– Takahiro Waki
Jul 31 at 14:57

@mrtaurho 1+1/e+1/e^2+・・・+1/e^n+・・・$=frac11-1/e=fracee-1$
– Takahiro Waki
Jul 31 at 14:57

 | 
show 1 more comment

up vote
7
down vote

We have that

$$fracsumlimits_k=1^n k^nn^n=sum_k=1^n left( frac k n right)^n=sum_k=0^n-1 left( frac n-k n right)^n=sum_k=0^n-1 left( 1-frac k n right)^n$$

then, following the suggestion given by ComplexYetTrivial, let consider

$$a_n=sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n$$

which is strictly increasing and bounded indeed by AM-GM we have

• $sqrt[n+1]left(1-fracknright)^n cdot 1 leq fracn
  left(1 - fracknright) + 1n+1 = 1 - frackn+1 $

and

• $left( 1-frac k n right)^n=e^nlog left( 1-frac k n
  right)=e^n left( -frac k n-frac k^2 2n^2-frac k^3
  3n^3-ldots right)le e^-k$

therefore by the monotone convergence theorem $a_n$ has finite limit and since for $k$ fixed

• $left( 1-frac k n right)^nto e^-k$

we have

$$lim_nto inftyfracsumlimits_k=1^n k^nn^n=lim_nto infty sum_k=0^n-1 left( 1-frac k n right)^n=lim_nto infty sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n=sum_k=0^inftye^-k=fracee-1$$

share|cite|improve this answer

edited Aug 1 at 21:08

answered Jul 31 at 14:46

gimusi

64.1k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) without Hints at last. :P
  – Mariusz Iwaniuk
  Jul 31 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk Thanks! The hint was already given! :)
  – gimusi
  Jul 31 at 15:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In your second line, the $O(frac1n)$ depends on $k$. Therefore you can’t factorize $O(frac1n)$ in the sum without an additional argument.
  – mathcounterexamples.net
  Jul 31 at 15:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net Yes I think you are right, it's a step I need to revise. Thanks
  – gimusi
  Jul 31 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could write the sum as $sum_k=0^infty left(1-fracknright)^n operatorname1(k<n)$, observe that the sequence $left[left(1-fracknright)^n operatorname1(k<n)right]_ninmathbbN$ is increasing and use the monotone convergence theorem to interchange summation and limit.
  – ComplexYetTrivial
  Jul 31 at 16:16
  
  

  
  
  
  

 | 
show 4 more comments

up vote
7
down vote

We have that

$$fracsumlimits_k=1^n k^nn^n=sum_k=1^n left( frac k n right)^n=sum_k=0^n-1 left( frac n-k n right)^n=sum_k=0^n-1 left( 1-frac k n right)^n$$

then, following the suggestion given by ComplexYetTrivial, let consider

$$a_n=sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n$$

which is strictly increasing and bounded indeed by AM-GM we have

• $sqrt[n+1]left(1-fracknright)^n cdot 1 leq fracn
  left(1 - fracknright) + 1n+1 = 1 - frackn+1 $

and

• $left( 1-frac k n right)^n=e^nlog left( 1-frac k n
  right)=e^n left( -frac k n-frac k^2 2n^2-frac k^3
  3n^3-ldots right)le e^-k$

therefore by the monotone convergence theorem $a_n$ has finite limit and since for $k$ fixed

• $left( 1-frac k n right)^nto e^-k$

we have

$$lim_nto inftyfracsumlimits_k=1^n k^nn^n=lim_nto infty sum_k=0^n-1 left( 1-frac k n right)^n=lim_nto infty sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n=sum_k=0^inftye^-k=fracee-1$$

share|cite|improve this answer

edited Aug 1 at 21:08

answered Jul 31 at 14:46

gimusi

64.1k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) without Hints at last. :P
  – Mariusz Iwaniuk
  Jul 31 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk Thanks! The hint was already given! :)
  – gimusi
  Jul 31 at 15:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In your second line, the $O(frac1n)$ depends on $k$. Therefore you can’t factorize $O(frac1n)$ in the sum without an additional argument.
  – mathcounterexamples.net
  Jul 31 at 15:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net Yes I think you are right, it's a step I need to revise. Thanks
  – gimusi
  Jul 31 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could write the sum as $sum_k=0^infty left(1-fracknright)^n operatorname1(k<n)$, observe that the sequence $left[left(1-fracknright)^n operatorname1(k<n)right]_ninmathbbN$ is increasing and use the monotone convergence theorem to interchange summation and limit.
  – ComplexYetTrivial
  Jul 31 at 16:16
  
  

  
  
  
  

 | 
show 4 more comments

up vote
7
down vote

up vote
7
down vote

We have that

$$fracsumlimits_k=1^n k^nn^n=sum_k=1^n left( frac k n right)^n=sum_k=0^n-1 left( frac n-k n right)^n=sum_k=0^n-1 left( 1-frac k n right)^n$$

then, following the suggestion given by ComplexYetTrivial, let consider

$$a_n=sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n$$

which is strictly increasing and bounded indeed by AM-GM we have

• $sqrt[n+1]left(1-fracknright)^n cdot 1 leq fracn
  left(1 - fracknright) + 1n+1 = 1 - frackn+1 $

and

• $left( 1-frac k n right)^n=e^nlog left( 1-frac k n
  right)=e^n left( -frac k n-frac k^2 2n^2-frac k^3
  3n^3-ldots right)le e^-k$

therefore by the monotone convergence theorem $a_n$ has finite limit and since for $k$ fixed

• $left( 1-frac k n right)^nto e^-k$

we have

$$lim_nto inftyfracsumlimits_k=1^n k^nn^n=lim_nto infty sum_k=0^n-1 left( 1-frac k n right)^n=lim_nto infty sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n=sum_k=0^inftye^-k=fracee-1$$

share|cite|improve this answer

edited Aug 1 at 21:08

answered Jul 31 at 14:46

gimusi

64.1k73480

We have that

$$fracsumlimits_k=1^n k^nn^n=sum_k=1^n left( frac k n right)^n=sum_k=0^n-1 left( frac n-k n right)^n=sum_k=0^n-1 left( 1-frac k n right)^n$$

then, following the suggestion given by ComplexYetTrivial, let consider

$$a_n=sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n$$

which is strictly increasing and bounded indeed by AM-GM we have

• $sqrt[n+1]left(1-fracknright)^n cdot 1 leq fracn
  left(1 - fracknright) + 1n+1 = 1 - frackn+1 $

and

• $left( 1-frac k n right)^n=e^nlog left( 1-frac k n
  right)=e^n left( -frac k n-frac k^2 2n^2-frac k^3
  3n^3-ldots right)le e^-k$

therefore by the monotone convergence theorem $a_n$ has finite limit and since for $k$ fixed

• $left( 1-frac k n right)^nto e^-k$

we have

$$lim_nto inftyfracsumlimits_k=1^n k^nn^n=lim_nto infty sum_k=0^n-1 left( 1-frac k n right)^n=lim_nto infty sum_substackk=0 \ k<n^infty left( 1-frac k n right)^n=sum_k=0^inftye^-k=fracee-1$$

share|cite|improve this answer

edited Aug 1 at 21:08

answered Jul 31 at 14:46

gimusi

64.1k73480

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 21:08

edited Aug 1 at 21:08

edited Aug 1 at 21:08

answered Jul 31 at 14:46

gimusi

64.1k73480

answered Jul 31 at 14:46

gimusi

64.1k73480

answered Jul 31 at 14:46

gimusi

64.1k73480

64.1k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) without Hints at last. :P
  – Mariusz Iwaniuk
  Jul 31 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk Thanks! The hint was already given! :)
  – gimusi
  Jul 31 at 15:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In your second line, the $O(frac1n)$ depends on $k$. Therefore you can’t factorize $O(frac1n)$ in the sum without an additional argument.
  – mathcounterexamples.net
  Jul 31 at 15:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net Yes I think you are right, it's a step I need to revise. Thanks
  – gimusi
  Jul 31 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could write the sum as $sum_k=0^infty left(1-fracknright)^n operatorname1(k<n)$, observe that the sequence $left[left(1-fracknright)^n operatorname1(k<n)right]_ninmathbbN$ is increasing and use the monotone convergence theorem to interchange summation and limit.
  – ComplexYetTrivial
  Jul 31 at 16:16
  
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) without Hints at last. :P
  – Mariusz Iwaniuk
  Jul 31 at 14:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk Thanks! The hint was already given! :)
  – gimusi
  Jul 31 at 15:01
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  In your second line, the $O(frac1n)$ depends on $k$. Therefore you can’t factorize $O(frac1n)$ in the sum without an additional argument.
  – mathcounterexamples.net
  Jul 31 at 15:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mathcounterexamples.net Yes I think you are right, it's a step I need to revise. Thanks
  – gimusi
  Jul 31 at 15:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could write the sum as $sum_k=0^infty left(1-fracknright)^n operatorname1(k<n)$, observe that the sequence $left[left(1-fracknright)^n operatorname1(k<n)right]_ninmathbbN$ is increasing and use the monotone convergence theorem to interchange summation and limit.
  – ComplexYetTrivial
  Jul 31 at 16:16
  
  

  
  
  
  

(+1) without Hints at last. :P
– Mariusz Iwaniuk
Jul 31 at 14:56

(+1) without Hints at last. :P
– Mariusz Iwaniuk
Jul 31 at 14:56

@MariuszIwaniuk Thanks! The hint was already given! :)
– gimusi
Jul 31 at 15:01

@MariuszIwaniuk Thanks! The hint was already given! :)
– gimusi
Jul 31 at 15:01

2

2

In your second line, the $O(frac1n)$ depends on $k$. Therefore you can’t factorize $O(frac1n)$ in the sum without an additional argument.
– mathcounterexamples.net
Jul 31 at 15:41

In your second line, the $O(frac1n)$ depends on $k$. Therefore you can’t factorize $O(frac1n)$ in the sum without an additional argument.
– mathcounterexamples.net
Jul 31 at 15:41

@mathcounterexamples.net Yes I think you are right, it's a step I need to revise. Thanks
– gimusi
Jul 31 at 15:44

@mathcounterexamples.net Yes I think you are right, it's a step I need to revise. Thanks
– gimusi
Jul 31 at 15:44

You could write the sum as $sum_k=0^infty left(1-fracknright)^n operatorname1(k<n)$, observe that the sequence $left[left(1-fracknright)^n operatorname1(k<n)right]_ninmathbbN$ is increasing and use the monotone convergence theorem to interchange summation and limit.
– ComplexYetTrivial
Jul 31 at 16:16

You could write the sum as $sum_k=0^infty left(1-fracknright)^n operatorname1(k<n)$, observe that the sequence $left[left(1-fracknright)^n operatorname1(k<n)right]_ninmathbbN$ is increasing and use the monotone convergence theorem to interchange summation and limit.
– ComplexYetTrivial
Jul 31 at 16:16

 | 
show 4 more comments

 

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