What does this series do: $\ln((2n+1)/(2n-1))$? (Converges or Diverges?)
1,120
Solution 1
The series diverges! To see why, look at a partial sum $$\sum_{n=1}^{N}{\log\left(\frac{2n+1}{2n-1}\right)} = \log\frac{3}{1} + \log\frac{5}{3} + \cdots + \log\frac{2N+1}{2N-1} \\ = \log\frac{2N+1}{1} \\ = \log(2N+1)$$ which diverges.
Solution 2
hint
$$\frac {2n+1}{2n-1}=1+\frac {2}{2n-1} $$
$$\ln (1+\frac {2}{2n-1})\sim \frac {2}{2n-1}\;\;(n\to+\infty) $$
$\sum \frac {2}{2n-1}$ diverges thus your series diverges as positive equivalent.
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Author by
MohammadReza Hosseini
Updated on January 02, 2020Comments
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MohammadReza Hosseini almost 4 years
what this series do when n is going to be a large number.
ln((2n+1)/(2n-1)) when n is tending to infinity?
I used integral test, ratio test, comparison test and ... what do you think about this series ???
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MohammadReza Hosseini over 6 yearshow the first line implies the second line: log (2N+1)/1?
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Li Chun Min over 6 yearsI believe that the limit of the $\ln((2x+1)/(2x-1))$ is zero
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Li Chun Min over 6 yearsUse log properties and you see things cancels out.
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lux over 6 yearsYou are quite right - I had not seen the factor of 2 in the denominator. Post edited.