Evaluate $\sum_{n=1}^\infty \frac1{n(2n-1)}$

Asked Feb 22 '18 at 19:43

Active Feb 22 '18 at 19:43

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The question is to evaluate $$\sum_{n=1}^\infty \frac1{n(2n-1)}$$

I have done this:

$$\sum_1^\infty\frac1{n(2n-1)}=\sum_{n=1}^\infty\frac2{2n-1}-\frac1n=\frac21-\frac11+\frac23-\frac12+\cdots\\=1-\frac12+\frac13-\frac14+\cdots=\log 2$$ But Wolfram alpha says the initial sum is $2\log 2$. I think the wrong step is when I started a new line above - the sum needs to be re-ordered to get it into the form I have got. But the series is not absolutely convergent (I think), so this shouldn't be allowed. Am I right? How can I correct this argument? If it is not possible to correct is, can someone provide an alternative method I could use instead?

asked Feb 22 '18 at 19:43

dialog

1

You are manipulating divergent series as though they were convergent. – Angina Seng Feb 22 '18 at 19:46
@LordSharktheUnknown Thanks! I couldn't find that question when I searched for it for some reason. I guess it is because I had a minus sign when I searched. – dialog Feb 22 '18 at 19:48
I think the error here is actually rearranging terms of a conditionally convergent series, which is not valid in general. – Daniel Schepler Feb 22 '18 at 19:51

Evaluate $\sum_{n=1}^\infty \frac1{n(2n-1)}$

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