What does this series do: $\ln((2n+1)/(2n-1))$? (Converges or Diverges?)

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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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MohammadReza Hosseini
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Updated on January 02, 2020

Comments

  • MohammadReza Hosseini
    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 ???

  • MohammadReza Hosseini
    MohammadReza Hosseini over 6 years
    how the first line implies the second line: log (2N+1)/1?
  • Li Chun Min
    Li Chun Min over 6 years
    I believe that the limit of the $\ln((2x+1)/(2x-1))$ is zero
  • Li Chun Min
    Li Chun Min over 6 years
    Use log properties and you see things cancels out.
  • lux
    lux over 6 years
    You are quite right - I had not seen the factor of 2 in the denominator. Post edited.