- #1

Rectifier

Gold Member

- 313

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**The problem**

I'd like to calculate the value of this sum:

$$3 \sum^\infty_{k=1}\frac{1}{2k^2-k}$$

**The attempt**

## 3 \sum^\infty_{k=1}\frac{1}{2k^2-k} = [k=t/2] = 3 \sum^\infty_{t=2}\frac{1}{2 \left( \frac{t}{2} \right)^2-\frac{t}{2}} = 3 \sum^\infty_{t=2}\frac{1}{ \frac{t^2}{2} - \frac{t}{2}} = 3 \sum^\infty_{t=2}\frac{2}{t^2-t} = \\ = 6 \sum^\infty_{t=2} \frac{1}{t(t-1)} = [partial \ factoring] = 6 \sum^\infty_{t=2} \frac{1}{t-1} - \frac{1}{t} = \\ = \lim_{n \rightarrow \infty} 6 \sum^n_{t=2}\frac{1}{t-1} - \frac{1}{t} =\\ = \lim_{n \rightarrow \infty} 6 \sum^n_{t=2} \left( \frac{1}{2-1}-\frac{1}{2} \right) + \left( \frac{1}{3-1}-\frac{1}{3} \right) + ... + \left( \frac{1}{(n-1)-1}-\frac{1}{n-1} \right) + \left( \frac{1}{n-1}-\frac{1}{n} \right) = \\ = \lim_{n \rightarrow \infty} 6 \sum^n_{t=2} \left( \frac{1}{1}-\frac{1}{2} \right) + \left( \frac{1}{2}-\frac{1}{3} \right) + ... + \left( \frac{1}{n-2}-\frac{1}{n-1} \right) + \left( \frac{1}{n-1}-\frac{1}{n} \right) = \\ = [telescoping \ series] = \lim_{n \rightarrow \infty} 6 \sum^n_{t=2} 1 -\frac{1}{n} = 6 ##

But I get the result 6ln(2) from

**wolfram**. Where did I make a mistake?