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$$\sum_{k=1}^{\infty}\ln(\frac{(k + 1) (k + 3)}{(k + 2) ^ 2})$$

The answer is = ln(2/3)

Can you please help me?

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  • $\begingroup$ Welcome to Mathematica StackExchange. This site is intended to discuss Wolfram Mathematica software. If you are looking for a mathematical, pen and paper explanation of how to calculate this sum, please refer to the Mathematics StackExchange. $\endgroup$
    – Domen
    Commented Mar 13, 2023 at 19:18
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    $\begingroup$ Worst title ever. Please rewrite to refer to the content of your question. Also, this is the wrong site for a math question unrelated to Mathematica software. $\endgroup$
    – David G. Stork
    Commented Mar 13, 2023 at 19:24
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    $\begingroup$ Sum[Log[(k + 1) (k + 3)/(k + 2)^2], {k, 1, Infinity}] does the job nicely. $\endgroup$
    – Daniel Huber
    Commented Mar 13, 2023 at 19:30
  • $\begingroup$ @DavidG.Stork Not sure about that - it is certainly honest. $\endgroup$
    – 1729taxi
    Commented Mar 13, 2023 at 20:22
  • $\begingroup$ Welcome to the Mathematica Stack Exchange. The introductory book written by the inventor is a good learning resource. There is a fast intro for math students as well as a fast intro for programmers to choose from. $\endgroup$
    – Syed
    Commented Mar 13, 2023 at 20:29

1 Answer 1

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$\begingroup$ 
Sum[Log[((k + 1)*(k + 3))/(k + 2)^2], {k, 1, Infinity}]

yields -Log[3/2] or Log[2/3] or Log[2]-Log[3].

That is what Wolfram Mathematica and WolframAlpha do.

Of course, I may have missed something.

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