There exists a bijective f:N→N such that:∑n=1∞(−1)f(n)ln⁡f(n)+1f(n)=ln⁡2010

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superpms01wks1 · Accepted Answer

Suppose first you take f(n)=2n then∑n=1∞(−1)2nln⁡2n+12n=∑n=1∞ln⁡(2n+1)−ln⁡(2n)=∑n=1∞12n+δn≥∑n=1∞12n+1=∞where 0&amp;lt;δn&amp;lt;1 - this is Lagrange&#039;s theorem f(x)−f(y)(x)−y}=f′(c) for some c∈(x,y), and here x=2n+1, y=2b, f(t)=ln⁡(t) so f′(t)=1t.The same works for f(n)=2n+1 but with −∞. Now construct another function f as follows : choose enough even numbers until the first time your partial sum is above ln⁡2100, then choose enough odd numbers until the first time your partial sum is below ln⁡2100. you can do this since the two sequences I described above converge to ±∞. you can continue like this to build a bijective map f:N→NSince the n−th term in the sum converges to zero, show that the sequence converges to ln⁡2100

There exists a bijective \(\displaystyle{f}:{\mathbb{{{N}}}}\to{\mathbb{{{N}}}}\) such that:

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