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

Question

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 ???

Answer 1

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.

Answer 2

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.

Answer 3

What happens to the terms in the series as n gets large? In the limit, the argument of the log tends to a constant value, call it c (you have to find it!)

Now that means you end up adding lots of copies of ln(c), infinitely many times. It should be obvious that this cannot converge to a finite answer.

Edit: in this case c=1 and the terms do tend to zero as pointed out in the comments