What is the sum of the following series?

The given series can be written as

$\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{4}{2n} - \dfrac{4}{2n + 1}\right) = 4 \times \left(\left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{4} - \dfrac{1}{5}\right) + \left(\dfrac{1}{6} - \dfrac{1}{7}\right) + ...\right) =$

$\displaystyle 4 \times \left(1 - \left(1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - .....\right)\right) = 4 \times \left(1 - \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}\right) = 4 \times (1 - \ln(2)) = \boxed{1.227}$

to 3 decimal places, where we recognize this last summation as that of the alternating harmonic series .

$\begin{aligned} S & = \lim_{m \to \infty} \sum_{n=1}^m \left(\frac 2n - \frac 4{2n+1} \right) \\ & = \lim_{m \to \infty} \left(\frac 21 - \frac 43 + \frac 22 - \frac 45 + \frac 23 - \frac 47 + ... + \frac 2m - \frac 4{2m+1}\right) \\ & = \lim_{m \to \infty} \left(2H_m - 4\left(H_{2m+1}-1-\frac 12H_m \right)\right) & \small \color{#3D99F6} \text{where }H_n \text{ is the }n \text{th harmonic number.} \\ & = \lim_{m \to \infty} \left(4 + 4H_m - 4H_{2m+1}\right) \\ & = \lim_{m \to \infty} 4 \left(1 + \ln m + {\color{#3D99F6} \gamma} - \ln (2m+1) - {\color{#3D99F6} \gamma} \right) & \small \color{#3D99F6} \text{where }\gamma \text{ is Euler-Mascheroni constant.} \\ & = \lim_{m \to \infty} 4 \left(1 + \ln \left(\frac m{2m+1} \right) \right) \\ & = 4 (1-\ln 2) \\ & \approx \boxed{1.227} \end{aligned}$

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References

• Harmonic number
• Euler-Mascheroni constant