Evaluate $\sum_{n=1}^\infty \ln\left( \frac{n(2n+1)}{(n+1)(2n-1)}\right)$

Question

Evaluate $\sum_{n=1}^\infty \ln\left( \frac{n(2n+1)}{(n+1)(2n-1)}\right)$

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My Approach

$$\sum_{n=1}^\infty \ln\left( \frac{n(2n+1)}{(n+1)(2n-1)}\right)=\sum_{n=1}^\infty ln(\left(n)(2n+1)\right) - ln(\left(n+1)(2n-1)\right) \\=\sum_{n=1}^\infty ln(n) + ln(2n+1) -ln(n+1)-ln(2n-1) $$

now we can re-write as $$\sum_{n=1}^\infty ln(n) -ln(n+1) \ + \sum_{n=1}^\infty ln(2n+1) - ln(2n-1)$$

Doing the partials sums we can find and take the limit of $S_n$ to find the value of the Series.

$$S_n = ln(1) + ln(1) -ln(n+1) +ln(2n+1) \\ \lim_{n \to \infty} S_n = \lim_{n \to \infty}ln\left(\frac{2n+1}{n+1}\right)= ln(2) $$

I want to know if what I did is legit because the answer is correct (according to the book)

Answer 1

$$\sum_{n=1}^\infty \ln(n) -\ln(n+1) \ + \sum_{n=1}^\infty \ln(2n+1) - \ln(2n-1)$$

Here you cannot split them into two summations, because both of them are divergent. Instead, you have to keep them into one summation,

$$\sum_{n=1}^\infty \left(\ln(n) -\ln(n+1) \right)+ (\ln(2n+1) - \ln(2n-1))$$

Next, do the parital sum as you did for $S_n$, then take the limit. Your result is correct.