Prove that $\sum_{n=1}^\infty \frac{2}{2n}-\frac{2}{2n+1}=2-2\ln(2)$

Question

The question is simple, to prove that $\sum_{n=1}^\infty{\frac{2}{2n}-\frac{2}{2n+1}}=2-2\ln(2)$.

I did it by wirtting down some terms of the series, rearranging as follows:

$$\sum_{n=1}^\infty \frac{2}{2n}-\frac{2}{2n+1}$$ $$2\sum_{n=1}^\infty \frac{1}{2n}-\frac{1}{2n+1}$$ $$2\left[\frac 12 - \frac 13 + \frac 14 - \frac 15+ \frac 16- \frac 17+ \frac 18- \frac 19+ \ldots\right]$$ $$2\left[1-\left(1-\frac 12 + \frac 13 -\frac 14 +\frac 15-\frac 16+\frac 17-\frac 18+\frac 19-\ldots\right)\right]$$

and noticing that the series inside the round parenthesis corresponds to $$1-\frac 12 + \frac 13 -\frac 14 +\frac 15-\frac 16+\frac 17-\frac 18+\frac 19-\ldots=\ln(2)$$ follows that:

$$2[1-\ln(2)]=2-2\ln(2)$$

So what I'm wanting to clarify is if is there any other way to improve this answer, a more "elegant" way to come up to the infinite sum of the series.

Answer 1

Well, you asked for an elegant way, which depends on your perspective for elegance. I came up with a solution using integrals. Simplifying the sum, $$ S =2\sum_{n=1}^{\infty} \frac{1}{2n}-\frac{1}{2n+1} = \sum_{n=1}^{\infty} \dfrac{1}{n(2n+1)} $$ Also, $$ \int_0^1 x^{2n}\ \mathrm{d} x = \dfrac{1}{2n+1} $$ Substituting this result into our sum, $$ S = \sum_{n=1}^{\infty} \int_0^1 \frac{x^{2n}}{n}\ \mathrm{d} x $$ Now, for $ |x| \lt 1 $, $$\ln(1-x)= -\sum_{n=1}^{\infty} \dfrac{x^n}{n} $$ Replacing $ x $ with $ x^2 $ $$ \ln(1-x^2)= -\sum_{n=1}^{\infty} \dfrac{x^{2n}}{n} $$ So, our sum will be transformed to $$ \begin{align*} S &= -\int_0^1 \ln(1-x^2)\ \mathrm{d} x \\ &= - \int_0^1 \ln(1-x)\ \mathrm{d} x -\int_0^1 \ln(1+x)\ \mathrm{d} x \end{align*} $$ Using the susbtitution $ 1-x \mapsto x $ on the first integral and $1+x \mapsto x $ on the second integral $$ \begin{align*} S &= -\int_0^1 \ln{x}\ \mathrm{d} x - \int_1^2 \ln(x)\ \mathrm{d} x \\ &= -\int_0^2 \ln{x}\ \mathrm{d} x \\ &= -\left. \biggl( x \ln{x}-x \biggr) \right|_0^2 \quad , \text{via IBP} \\ &= 2-2\ln{2} \end{align*} $$

Answer 2

Use $\frac{1}{m}=\int_{0}^{1} t^{m-} dt$ and sum of an infinite GP to get $$S=2\sum_{n=1}^\infty \left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\sum_{n=1}^\infty \left(\int_{0}^{1}( t^{2n-1}-t^{2n})dt\right)=2\int_{0}^{1} \left( \frac{t^{-1} t^2}{1-t^2}-\frac{t^2 }{1-t^2}\right)dt$$ $$=2\int_{0}^{1} \left(\frac{t}{1+t}\right)dt=2[1-\ln 2].$$