Value of an infinite sum

  • #1
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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?
 

Answers and Replies

  • #2
Orodruin
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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.
 
  • #3
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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}} ##
 
  • #4
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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.
 
  • #5
Ray Vickson
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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.
 
Last edited:

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