Is it possible to evaluate the summation $x=sum_n=1^infty fracphi(n)n^2$?Prove that $ sum_1 le t le n, (t, n) = 1 t = dfrac nphi(n)2 $Is the set $phi(mathbbN)$ syndetic?Is $a_n=frac1nsum_k=1^nfracvarphi(k)k$ convergent?What's equal this :$sum_phi(n)=1^inftyfrac1phi(n)$?What's the closed form of this :$sum_n=1^+inftyfrac(-1)^nphi(n)n$Prove that $phi(phi(n)) le fracn2$.On the series $sum_n=1^infty fracleft lfloor log_2 n right rfloor2n(2n+1)(2n+2)$Evaluating $sum_n=1^infty fracphi(n)7^n + 1$, where $phi(n)$ is Euler's totient functionIs the equation $x = phi(n), x=2k, n,k in mathbbZ$, where $phi(n)$ is the Euler totient function, solvable for all evens?Are these limits correct? $lim_nto inftytextsup fraclambda (n)n=1$ and $lim_nto inftytextinf fraclambda (n)n=0$ exist?
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Is it possible to evaluate the summation $x=sum_n=1^infty fracphi(n)n^2$?
Prove that $ sum_1 le t le n, (t, n) = 1 t = dfrac nphi(n)2 $Is the set $phi(mathbbN)$ syndetic?Is $a_n=frac1nsum_k=1^nfracvarphi(k)k$ convergent?What's equal this :$sum_phi(n)=1^inftyfrac1phi(n)$?What's the closed form of this :$sum_n=1^+inftyfrac(-1)^nphi(n)n$Prove that $phi(phi(n)) le fracn2$.On the series $sum_n=1^infty fracleft lfloor log_2 n right rfloor2n(2n+1)(2n+2)$Evaluating $sum_n=1^infty fracphi(n)7^n + 1$, where $phi(n)$ is Euler's totient functionIs the equation $x = phi(n), x=2k, n,k in mathbbZ$, where $phi(n)$ is the Euler totient function, solvable for all evens?Are these limits correct? $lim_nto inftytextsup fraclambda (n)n=1$ and $lim_nto inftytextinf fraclambda (n)n=0$ exist?
$begingroup$
$$x=sum_n=1^infty fracphi(n)n^a,quadtextwhere $phi$ is the Euler-phi/totient function and $ageq1$$$
Can this even be evaluated? It clearly converges for all $a>2$, since the maximum value of the totient function for any $n$ is $n-1$ and $x=sum_n=1^infty fracn-1n^a$ converges if $a>2$.
I honestly have no idea where to begin on this one.
sequences-and-series summation totient-function
asked Dec 18 '18 at 0:41
R. BurtonR. Burton
732110
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|
show 2 more comments
$begingroup$
$$x=sum_n=1^infty fracphi(n)n^a,quadtextwhere $phi$ is the Euler-phi/totient function and $ageq1$$$
Can this even be evaluated? It clearly converges for all $a>2$, since the maximum value of the totient function for any $n$ is $n-1$ and $x=sum_n=1^infty fracn-1n^a$ converges if $a>2$.
I honestly have no idea where to begin on this one.
sequences-and-series summation totient-function
asked Dec 18 '18 at 0:41
R. BurtonR. Burton
732110
$endgroup$
$begingroup$
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$, hence the series with $a=2$ diverges.
$endgroup$
– Kemono Chen
Dec 18 '18 at 0:43
$begingroup$
Ah, thank you! Um, doesn't $fraczeta(a-1)zeta(a)$ converge for real $a>2$? Also do know if there's a specific reference for this identity somewhere?
$endgroup$
– R. Burton
Dec 18 '18 at 0:51
$begingroup$
@KemonoChen the OP has asked for a reference, but neglected to put an at sign and your name
$endgroup$
– Will Jagy
Dec 18 '18 at 1:14
2
$begingroup$
I saw this result in Chapter 3 Exercise 6 and 8 of the book "Introduction to Analytic Number theory" by Tom M. Apostol.
$endgroup$
– Kemono Chen
Dec 18 '18 at 1:19
1
$begingroup$
It follows from $n = sum_d phi(d) $
$endgroup$
– reuns
Dec 18 '18 at 1:23
|
show 2 more comments
$begingroup$
$$x=sum_n=1^infty fracphi(n)n^a,quadtextwhere $phi$ is the Euler-phi/totient function and $ageq1$$$
Can this even be evaluated? It clearly converges for all $a>2$, since the maximum value of the totient function for any $n$ is $n-1$ and $x=sum_n=1^infty fracn-1n^a$ converges if $a>2$.
I honestly have no idea where to begin on this one.
sequences-and-series summation totient-function
asked Dec 18 '18 at 0:41
R. BurtonR. Burton
732110
$endgroup$
$$x=sum_n=1^infty fracphi(n)n^a,quadtextwhere $phi$ is the Euler-phi/totient function and $ageq1$$$
Can this even be evaluated? It clearly converges for all $a>2$, since the maximum value of the totient function for any $n$ is $n-1$ and $x=sum_n=1^infty fracn-1n^a$ converges if $a>2$.
I honestly have no idea where to begin on this one.
sequences-and-series summation totient-function
sequences-and-series summation totient-function
asked Dec 18 '18 at 0:41
R. BurtonR. Burton
732110
asked Dec 18 '18 at 0:41
R. BurtonR. Burton
732110
asked Dec 18 '18 at 0:41
R. BurtonR. Burton
732110
asked Dec 18 '18 at 0:41
R. BurtonR. Burton
732110
asked Dec 18 '18 at 0:41
R. BurtonR. Burton
732110
732110
$begingroup$
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$, hence the series with $a=2$ diverges.
$endgroup$
– Kemono Chen
Dec 18 '18 at 0:43
$begingroup$
Ah, thank you! Um, doesn't $fraczeta(a-1)zeta(a)$ converge for real $a>2$? Also do know if there's a specific reference for this identity somewhere?
$endgroup$
– R. Burton
Dec 18 '18 at 0:51
$begingroup$
@KemonoChen the OP has asked for a reference, but neglected to put an at sign and your name
$endgroup$
– Will Jagy
Dec 18 '18 at 1:14
2
$begingroup$
I saw this result in Chapter 3 Exercise 6 and 8 of the book "Introduction to Analytic Number theory" by Tom M. Apostol.
$endgroup$
– Kemono Chen
Dec 18 '18 at 1:19
1
$begingroup$
It follows from $n = sum_d phi(d) $
$endgroup$
– reuns
Dec 18 '18 at 1:23
|
show 2 more comments
$begingroup$
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$, hence the series with $a=2$ diverges.
$endgroup$
– Kemono Chen
Dec 18 '18 at 0:43
$begingroup$
Ah, thank you! Um, doesn't $fraczeta(a-1)zeta(a)$ converge for real $a>2$? Also do know if there's a specific reference for this identity somewhere?
$endgroup$
– R. Burton
Dec 18 '18 at 0:51
$begingroup$
@KemonoChen the OP has asked for a reference, but neglected to put an at sign and your name
$endgroup$
– Will Jagy
Dec 18 '18 at 1:14
2
$begingroup$
I saw this result in Chapter 3 Exercise 6 and 8 of the book "Introduction to Analytic Number theory" by Tom M. Apostol.
$endgroup$
– Kemono Chen
Dec 18 '18 at 1:19
1
$begingroup$
It follows from $n = sum_d phi(d) $
$endgroup$
– reuns
Dec 18 '18 at 1:23
$begingroup$
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$, hence the series with $a=2$ diverges.
$endgroup$
– Kemono Chen
Dec 18 '18 at 0:43
$begingroup$
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$, hence the series with $a=2$ diverges.
$endgroup$
– Kemono Chen
Dec 18 '18 at 0:43
$begingroup$
Ah, thank you! Um, doesn't $fraczeta(a-1)zeta(a)$ converge for real $a>2$? Also do know if there's a specific reference for this identity somewhere?
$endgroup$
– R. Burton
Dec 18 '18 at 0:51
$begingroup$
Ah, thank you! Um, doesn't $fraczeta(a-1)zeta(a)$ converge for real $a>2$? Also do know if there's a specific reference for this identity somewhere?
$endgroup$
– R. Burton
Dec 18 '18 at 0:51
$begingroup$
@KemonoChen the OP has asked for a reference, but neglected to put an at sign and your name
$endgroup$
– Will Jagy
Dec 18 '18 at 1:14
$begingroup$
@KemonoChen the OP has asked for a reference, but neglected to put an at sign and your name
$endgroup$
– Will Jagy
Dec 18 '18 at 1:14
2
2
$begingroup$
I saw this result in Chapter 3 Exercise 6 and 8 of the book "Introduction to Analytic Number theory" by Tom M. Apostol.
$endgroup$
– Kemono Chen
Dec 18 '18 at 1:19
$begingroup$
I saw this result in Chapter 3 Exercise 6 and 8 of the book "Introduction to Analytic Number theory" by Tom M. Apostol.
$endgroup$
– Kemono Chen
Dec 18 '18 at 1:19
1
1
$begingroup$
It follows from $n = sum_d phi(d) $
$endgroup$
– reuns
Dec 18 '18 at 1:23
$begingroup$
It follows from $n = sum_d phi(d) $
$endgroup$
– reuns
Dec 18 '18 at 1:23
|
show 2 more comments
1 Answer
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oldest
votes
$begingroup$
From Kemono Chen: This series is discussed in Chapter 3 of Tom M. Apostol's Introduction to Analytic Number Theory. Excercises 6 and 8 provide the result
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$
With the implication that the series diverges for $a=2$
From Slade: While the sum is finite for $a>2$, the limit as $ato2^+$ is $+infty$.
answered Mar 21 at 22:48
R. BurtonR. Burton
732110
$endgroup$
add a comment |
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$begingroup$
From Kemono Chen: This series is discussed in Chapter 3 of Tom M. Apostol's Introduction to Analytic Number Theory. Excercises 6 and 8 provide the result
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$
With the implication that the series diverges for $a=2$
From Slade: While the sum is finite for $a>2$, the limit as $ato2^+$ is $+infty$.
answered Mar 21 at 22:48
R. BurtonR. Burton
732110
$endgroup$
add a comment |
$begingroup$
From Kemono Chen: This series is discussed in Chapter 3 of Tom M. Apostol's Introduction to Analytic Number Theory. Excercises 6 and 8 provide the result
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$
With the implication that the series diverges for $a=2$
From Slade: While the sum is finite for $a>2$, the limit as $ato2^+$ is $+infty$.
answered Mar 21 at 22:48
R. BurtonR. Burton
732110
$endgroup$
add a comment |
$begingroup$
From Kemono Chen: This series is discussed in Chapter 3 of Tom M. Apostol's Introduction to Analytic Number Theory. Excercises 6 and 8 provide the result
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$
With the implication that the series diverges for $a=2$
From Slade: While the sum is finite for $a>2$, the limit as $ato2^+$ is $+infty$.
answered Mar 21 at 22:48
R. BurtonR. Burton
732110
$endgroup$
From Kemono Chen: This series is discussed in Chapter 3 of Tom M. Apostol's Introduction to Analytic Number Theory. Excercises 6 and 8 provide the result
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$
With the implication that the series diverges for $a=2$
From Slade: While the sum is finite for $a>2$, the limit as $ato2^+$ is $+infty$.
answered Mar 21 at 22:48
R. BurtonR. Burton
732110
answered Mar 21 at 22:48
R. BurtonR. Burton
732110
answered Mar 21 at 22:48
R. BurtonR. Burton
732110
answered Mar 21 at 22:48
R. BurtonR. Burton
732110
732110
add a comment |
add a comment |
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$begingroup$
$$sum_n=1^infty fracphi(n)n^a=fraczeta(a-1)zeta(a)$$, hence the series with $a=2$ diverges.
$endgroup$
– Kemono Chen
Dec 18 '18 at 0:43
$begingroup$
Ah, thank you! Um, doesn't $fraczeta(a-1)zeta(a)$ converge for real $a>2$? Also do know if there's a specific reference for this identity somewhere?
$endgroup$
– R. Burton
Dec 18 '18 at 0:51
$begingroup$
@KemonoChen the OP has asked for a reference, but neglected to put an at sign and your name
$endgroup$
– Will Jagy
Dec 18 '18 at 1:14
2
$begingroup$
I saw this result in Chapter 3 Exercise 6 and 8 of the book "Introduction to Analytic Number theory" by Tom M. Apostol.
$endgroup$
– Kemono Chen
Dec 18 '18 at 1:19
1
$begingroup$
It follows from $n = sum_d phi(d) $
$endgroup$
– reuns
Dec 18 '18 at 1:23