How can I find the sum of $\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}$

Question

How can I find the sum of the series:

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

I did this:

$$\sum_{n=2}^\infty \left(\frac{(-1)^n}{n-1} + \frac{(-1)^{n-1}}{n}\right)$$

But I am not sure how to proceed.

Answer 1

You are on the right track: $$\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}=\sum_{n=2}^\infty \left(\frac{(-1)^n}{n-1} + \frac{(-1)^{n-1}}{n}\right).$$ Now note that $$\ln(2)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n-1}.$$

Answer 2

$$ \sum_{n=2}^\infty \frac{(-1)^n}{n^2-n}= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}=\lim_{x\to 1^{-}} \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{n+1}}{n(n+1)} =\lim_{x\to 1^{-}}\int_0^x\sum_{n=1}^\infty \frac{(-1)^{n+1}t^n}{n}\,dt \\=\lim_{x\to 1^{-}}\int_0^x \log (t+1)\,dt=\lim_{x\to 1^{-}}\big((x+1)\log (x+1)-x\big)=2\log 2-1 $$ Note. The second "=" is a consequence of Abel's Theorem.

Answer 3

Let us consider a sequence $$a_n = \sum_{k=n+1}^{2n} \frac{1}{k}$$ Now, let's attempt to get a recursive formula for this. $$a_n - a_{n-1} = \sum_{k=n+1}^{2n} \frac{1}{k} - \sum_{k=n}^{2n-2} \frac{1}{k} = -\frac{1}{n} + \frac{1}{2n-1} + \frac{1}{2n} = \frac{1}{(2n-1)(2n)}$$ $$a_n = a_{n-1} + \frac{1}{(2n-1)(2n)}$$ Since $a_1 = \frac{1}{2} = \frac{1}{(2\cdot1 - 1)(2\cdot1)}$, $$a_n = \sum_{k=1}^n \frac{1}{(2k-1)(2k)}$$ Now if we also consider the sequence alternating harmonic series $$S_n = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{(-1)^{n-1}}{n}$$ We can see that $$S_{2n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n - 1} - \frac{1}{2n}$$ $$S_{2n} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{2n - 1} - \frac{1}{2n}\right)$$ $$S_{2n} = \sum_{k=1}^n \left(\frac{1}{2k-1} - \frac{1}{2k}\right) = \sum_{k=1}^n \frac{1}{(2k-1)(2k)}$$ This means that $S_{2n} = a_n$. If we now look at the expansion of $a_n$ $$a_n = \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \cdots + \frac{1}{2n}$$ If we factor out $\frac{1}{n}$, we get $$a_n = \frac{1}{n} \left(\frac{1}{1+\frac{1}{n}} + \frac{1}{1+\frac{2}{n}} + \frac{1}{1+\frac{3}{n}} + \cdots + \frac{1}{1+\frac{n}{n}}\right)$$ As $n$ tends towards infinity, this is the Riemann sum for $\int_0^1 \frac{1}{1+x}~{\rm d}x$. $$\int_0^1 \frac{1}{1+x}~{\rm d}x = \int_1^2 \frac{1}{x}~{\rm d}x = \ln 2$$ You can then use that to solve your sum.

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Alternative: Consider successive derivatives of $\ln(1 + x)$. $$\frac{{\rm d}}{{\rm d}x} \left[\ln(1 + x)\right] = \frac{1}{1 + x} = \left(1 + x\right)^{-1}$$ $$\frac{{\rm d}^2}{{\rm d}x^2} \left[\ln(1 + x)\right] = -\left(1 + x\right)^{-2}$$ $$\frac{{\rm d}^3}{{\rm d}x^3} \left[\ln(1 + x)\right] = 2\left(1 + x\right)^{-3}$$ $$\frac{{\rm d}^4}{{\rm d}x^4} \left[\ln(1 + x)\right] = -6\left(1 + x\right)^{-4}$$ $$\vdots$$ $$\frac{{\rm d}^n}{{\rm d}x^n} \left[\ln(1 + x)\right] = (-1)^{n-1} \cdot (n-1)! \cdot \left(1 + x\right)^{-n}$$ Using this to create a Taylor series about $x=0$ gives us $$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots + \frac{(-1)^{n-1} \cdot x^n}{n} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \cdot x^n}{n}$$ Plugging in $x=1$ to give the alternating harmonic series gives us $$\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{(-1)^{n-1}}{n} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$$

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Now knowing that the alternating harmonic series is equal to $\ln 2$, we can solve the sum. $$\sum_{n=2}^\infty \left(\frac{(-1)^n}{n-1} + \frac{(-1)^{n-1}}{n}\right)$$ $$= \sum_{n=2}^\infty \frac{(-1)^n}{n-1} + \sum_{n=2}^\infty \frac{(-1)^{n-1}}{n}$$ $$= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} + \left(-1 + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\right)$$ $$= \ln (2) + \ln (2) - 1 = \ln (4) - 1$$