What Is the Value of the Infinite Sum \(3 \sum_{k=1}^\infty \frac{1}{2k^2-k}\)?

[Rectifier]

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?

 

[Orodruin]

When you put k = t/2 without being careful about the summation set ... since k is an integer you should only sum over even t.

 

[Rectifier]

May I ask how I can bypass that error?

$3 \sum^\infty_{k=1}\frac{1}{2k^2-k} = [k=t/2] = 3 \sum^\infty_{t=2}\frac{1}{2 \left( \frac{t}{2} \right)^2-\frac{t}{2}}$

 

[fresh_42]

May I ask how I can bypass that error?

$3 \sum^\infty_{k=1}\frac{1}{2k^2-k} = [k=t/2] = 3 \sum^\infty_{t=2}\frac{1}{2 \left( \frac{t}{2} \right)^2-\frac{t}{2}}$

You confused the counting width here, but you do not need the substitution. The partial factoring works without. Then write down a few terms and you should see what the sum is.

 

[Ray Vickson]

May I ask how I can bypass that error?

$3 \sum^\infty_{k=1}\frac{1}{2k^2-k} = [k=t/2] = 3 \sum^\infty_{t=2}\frac{1}{2 \left( \frac{t}{2} \right)^2-\frac{t}{2}}$

Let $H(N) = \sum_{n=1}^N 1/n$ be the Harmonic function; see, eg.,
https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Let $$S(N) = \sum_{k=1}^N 1/(2k^2-k).$$
Expand $1/(2k^2-k)$ into partial fractions and note that one of the parts of $S(N)$ involves the sum of the odd reciprocals from $1$ to $2N-1$, that is $S_{\text{odd}}$:
$$S_{\text{odd}} =\sum_{k=1}^N 1/(2k-1).$$
This is the sum of all reciprocals from 1 to $2N$, less the sum of the even reciprocals from 2 to $2N$; the latter can be expressed in terms of $H(N)$. Thus, we can express $S(N)$ exactly in terms of $H(N)$ and $H(2N)$. The limit as $N \to \infty$ is then do-able.