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n → ∞ lim ζ ( 2 n ) ( ζ ( n ) ) 2 = ?
Notation : ζ ( ⋅ ) denotes the Riemann zeta function .
The answer is 1.0.
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2 solutions
Moderator note:
What is a simple way of showing that the limit is 1? Clearly it is bounded below by 1.
for Challenge Master: I remember us doing a problem last year were we showed that ∑ n = 2 ∞ ( ζ ( n ) − 1 ) = 1 ; that will do it.
Perfect solution as always Otto! (+1)
How did you get the long horizontal bar on the previous problem? Is there a latex command for it?
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No, the good people from Brilliant have added that ;)
after writing your stuff give one line space and type without any brackets or spaces
(_ _ _)
works fine
cool right?
this can be written as n → ∞ lim ∏ p = p r i m e 1 − p − 2 n 1 ( ∏ p = p r i m e 1 − p − n 1 ) 2 using properties of products: n → ∞ lim p = p r i m e ∏ 1 − p − 2 n 1 ( 1 − p − n 1 ) 2 using differences of 2 squares identity: n → ∞ lim p = p r i m e ∏ ( 1 − p − n 1 ) ( 1 + p − n 1 ) ( 1 − p − n 1 ) 2 = n → ∞ lim p = p r i m e ∏ 1 + p − n 1 1 − p − n 1 this is easily writeable as n → ∞ lim p = p r i m e ∏ p n − 1 p n + 1 = n → ∞ lim p = p r i m e ∏ ( 1 + p n − 1 2 ) as n aproaches infinity, so does p n − 1 . and hence p n − 1 2 becomes zero. the limit becomes p = p r i m e ∏ ( 1 ) = 1 note: the euler product is bieng used here.
Your solution actually doesn't work. Be careful when you are switching the order of limits. The "remaining parts" are actually very important to deal with, and even though these remainder may go to zero, the product could still be infinite. For example, consider
n → ∞ lim m = 1 ∏ ∞ ( 1 + m + n 1 )
If we take the product first, we obtain the telescoping series - product which equals to infinity, and hence the limit to infinity is infinity. However, if we were to interchange the limits and evaluate as n → ∞ , we will obtain the product of 1, which is 1.
For an alternative interpretation, take logarithms, which converts this to the standard summation approach.
nice problem!
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Thank you, it's a nice solution too.
Although there is a really short solution (see Otto's solution). It's the reason i put this problem as a level 3 problem rather than 4 or higher.
lim n → ∞ ζ ( n ) = 1 so the limit we seek is 1 as well by the basic limit rules.