How does one prove $\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)(n+2)} = 2 \log 2 -1$?

Question

$$\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)(n+2)} = 2 \log 2 -1$$ Saw this result on wikipedia, but I can’t seem to find a proof of this anywhere.

I’m rather slow, so if you know the solution, I’d appreciate it if it were worked out, instead of merely hinted at.

Answer 1

The sum can be found using our favourite alternative method of converting the sum to a double integral first.

Noting that $$\frac{1}{n + 1} = \int_0^1 x^n \, dx \quad \text{and} \quad \frac{1}{n + 2} = \int_0^1 y^{n + 1} \, dy,$$ the sum can be rewritten as \begin{align*} \sum_{n = 0}^\infty \frac{(-1)^n}{(n + 1)(n + 2)} &= \sum_{n = 0}^\infty (-1)^n \int_0^1 \int_0^1 x^n y^{n + 1} \, dx dy\\ &= \int_0^1 \int_0^1 y \sum_{n = 0}^\infty (-xy)^n dx dy \tag1\\ &= \int_0^1 \int_0^1 y \cdot \frac{1}{1 + xy} \, dx dy \tag2\\ &= \int_0^1 \ln (1 + xy) \Big{|}_0^1 \, dy\\ &= \int_0^1 \ln (1 + y) \, dy\\ &= \Big{[}(1 + y) \ln (1 + y) - (1 + y) \Big{]}_0^1\\ &= 2 \ln (2) - 1, \end{align*} as required.

Explanation

(1) Interchanging the summation with the double integration.

(2) Summing the series which is geometric.

Answer 2

HINT

Note that

$$\frac{1}{(n+1)(n+2)}=\frac{1}{(n+1)}-\frac{1}{(n+2)}$$

and Alternating harmonic series convergence to $\log 2$.

Answer 3

Ok, let me offer a terrible 'proof' based on hypergeometric functions. Consider the series:

$$f(x)=\sum_{n=0}^\infty \frac{x^n}{(n+1)(n+2)} $$

Let's check the ratio of subsequent general terms (see Wikipedia):

$$\frac{t_{n+1}}{t_n}=\frac{(n+1)(n+2)}{(n+2)(n+3)}x=\frac{(n+1)}{(n+3)}x=\frac{(n+1)(n+1)}{(n+3)}\frac{x}{n+1}$$

Which by definition makes the series proportional to the hypergeometric function:

$$f(x)= \frac{1}{2} {_2 F_1} (1,1;3;x)$$

The factor before the function is $t_0=1/2$.

The general form of this function is (you can check with Wolfram Alpha for example, or you can use another definition of the function, such as the integral definition):

$${_2 F_1} (1,1;3;x)=2 \frac{x+(1-x) \log (1-x)}{x^2}$$

Substituting $x=-1$, we obtain:

$$f(-1)= \frac{1}{2} {_2 F_1} (1,1;3;-1)=2 \log 2-1$$

This 'proof' is just for fun, but getting to hypergeometric form can be a valid method for a lot of other series, because there's a lot of reduction formulas for hypergeometric functions, and programs such as Mathematica can easily reduce them.