Zeta pushed to the limit 3

Calculus Level 5

lim ⁡ n → ∞ ζ ′ ( H n ) ζ ′ ( ln ⁡ n ) \lim_{n \to \infty} \frac{\zeta'(H_n)}{\zeta'(\ln n)}

Evaluate the limit above, where ζ ′ ( s ) = d d s ζ ( s ) \zeta'(s) = \dfrac{d}{ds} \zeta(s) .

You might also enjoy part 1 and/or 2 .


The answer is 0.67025609338.

Jake Lai by Jake Lai , 21, Singapore

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1 solution

Julian Poon
Nov 5, 2015

ζ ′ ( s ) = − ∑ k = 1 ∞ ln ⁡ ( k ) k s \zeta '(s)=-\sum _{ k=1 }^{ \infty }{ \frac { \ln { (k) } }{ { k }^{ s } } }

Since as n n approaches infinity, H n = k + ln ⁡ n H_{n}=k+\ln{n} , where k k is the Euler–Mascheroni constant, lim ⁡ n → ∞ ζ ′ ( H n ) ζ ′ ( ln ⁡ n ) = lim ⁡ n → ∞ ζ ′ ( ln ⁡ n + k ) ζ ′ ( ln ⁡ n ) = lim ⁡ n → ∞ ζ ′ ( n + k ) ζ ′ ( n ) \lim _{ n\to \infty } \frac { \zeta '(H_{ n }) }{ \zeta '(\ln { n } ) } =\lim _{ n\to \infty } \frac { \zeta '(\ln { n } +k) }{ \zeta '(\ln { n } ) } =\lim _{ n\to \infty } \frac { \zeta '(n+k) }{ \zeta '(n) }

= ln ⁡ ( 2 ) 2 n + k + ln ⁡ ( 3 ) 3 n + k . . . ln ⁡ ( 2 ) 2 n + ln ⁡ ( 3 ) 3 n . . . = ln ⁡ ( 2 ) + ln ⁡ ( 3 ) 2 n + k 3 n + k . . . ln ⁡ ( 2 ) 2 k + ln ⁡ ( 3 ) 2 n + k 3 n . . . = 2 − k =\frac { \frac { \ln { \left( 2 \right) } }{ 2^{ n+k } } +\frac { \ln { \left( 3 \right) } }{ 3^{ n+k } } ... }{ \frac { \ln { \left( 2 \right) } }{ 2^{ n } } +\frac { \ln { \left( 3 \right) } }{ 3^{ n } } ... } =\frac { \ln { \left( 2 \right) } +\frac { \ln { \left( 3 \right) } 2^{ n+k } }{ 3^{ n+k } } ... }{ \ln { \left( 2 \right) } 2^{ k }+\frac { \ln { \left( 3 \right) } 2^{ n+k } }{ 3^{ n } } ... } =2^{ -k }

Hence lim ⁡ n → ∞ ζ ′ ( H n ) ζ ′ ( ln ⁡ n ) = 2 − k = 0.6702560933... \lim _{ n\to \infty } \frac { \zeta '(H_{ n }) }{ \zeta '(\ln { n } ) } =2^{ -k }=\boxed{0.6702560933...}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

There are several issues with this solution:

  1. It is not true that H n = k + ln ⁡ n H_n = k + \ln n . What we have is that \lim_ H_n - \ln n = k .
  2. For a continuous function f f , and two sequences a n , b n a_n, b_n with lim ⁡ a n − b n = 0 \lim a_n - b_n = 0 , it is not true that lim ⁡ f ( a n ) − f ( b n ) = 0 \lim f(a_n) - f(b_n) = 0 . It is true in the case where lim ⁡ a n \lim a_n is finite, but with a n = H n a_n = H_n , the limit is infinity. As a simple counter example, consider a n = n , b n = n + 1 n a_n = n, b_n = n + \frac{1}{n} and f ( x ) = x 2 f(x) = x^2 . We have lim ⁡ f ( a n ) − f ( b n ) = 2 \lim f(a_n) - f(b_n) = 2 .

As such, the second line hasn't been justified.

Actually, ζ ′ ( s ) = − ∑ n = 1 ∞ ln ⁡ n n s \displaystyle \zeta'(s) = -\sum_{n=1}^{\infty} \frac{\ln n}{n^s} , so you're missing a negative sign.

Alternatively, one can show that lim ⁡ ζ ′ ( a ) ζ ′ ( b ) = lim ⁡ ζ ( a ) − 1 ζ ( b ) − 1 \lim \frac{\zeta'(a)}{\zeta'(b)} = \lim \frac{\zeta(a)-1}{\zeta(b)-1} by l'Hopital's rule.

Jake Lai - 5 years, 7 months ago

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I too realised that and made a problem on it.

Julian Poon - 5 years, 7 months ago

Issue 1: Since in the question, we are considering lim ⁡ n → ∞ H n \lim_{n\to \infty} H_{n} , it should be okay to use lim ⁡ n → ∞ H n = ln ⁡ ( n ) + k \lim_{n\to \infty} H_{n}=\ln(n)+k .

Issue 2: Yeah, that part is wrong. Opps. The answer is still 2 − k 2^{-k} though since in the later steps, sticking to ln ⁡ ( n ) \ln(n) still makes the limit converge to 2 − k 2^{-k} .

Julian Poon - 5 years, 6 months ago

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