Series sum by Riemann’s zeta function

(FB/Math/BM – Proposed by Souradeep Purkayastha)

Find the series sum  ln2/2 – ln3/3 + ln4/4 – ln5/5 + ….

___________________________________________________

Define, for x < 0,

f(x) = 2ˣ·ln2/2 – 3ˣ·ln3/3 + 4ˣ·ln4/4 – 5ˣ·ln5/5 + … =

= d/dx (2ˣ/2 – 3ˣ/3 + 4ˣ/4 – 5ˣ/5 + …) =

= d/dx [1 – (1ˣ⁻¹ – 2ˣ⁻¹ + 3ˣ⁻¹ – 4ˣ⁻¹ + 5ˣ⁻¹ – …)] =

= d/dx [1 – (1 – 2ˣ)·zeta(1 – x)] =
(*)
= d/dx [1 – (1 – 2ˣ)·(-1/x + γ + γ₁x + O(x²))] =

= 2ˣ·ln 2·(-1/x + γ + γ₁x + O(x²)) – (1 – 2ˣ)·(1/x² + γ₁x + O(x)) =

= 2ˣ·ln 2·(-1/x) – (1 – 2ˣ)·(1/x²) + 2ˣ·ln 2·γ + O(x)

Then ln 2/2 – ln 3/3 + ln 4/4 – ln 5/5 + … = lim [x→0-] f(x)
——————-
The first term:

2ˣ·ln 2·(-1/x) – (1 – 2ˣ)·(1/x²) =

= [2ˣ·ln 2·(-x) – (1 – 2ˣ)]·1/x² =

= [2ˣ·(1 -x·ln 2) – 1]·1/x²

The first two derivatives of this are

2ˣ·ln2·(1 -x·ln 2) – 2ˣ·ln2 = -x·2ˣ·ln²2,
-2ˣ·ln²2 – x·2ˣ·ln³2

so, by l’Hospital’s rule

lim [x→0-] 2ˣ·ln 2·(-1/x) – (1 – 2ˣ)·(1/x²) = -ln²2/2 = -ln 2
——————–
Then all together:

lim [x→0-] f(x) = -ln 2 + ln 2·γ
———————————————————
ln 2/2 – ln 3/3 + ln 4/4 – ln 5/5 + … = (γ – 1)ln2
———————————————————

(*)
zeta(s) = 1/(s – 1) + γ – γ₁(s – 1) + O((s – 1)²), near s = 0
zeta(1 – x) = -1/x + γ + γ₁x + O(x²), near x = 1

http://mathworld.wolfram.com/RiemannZetaFunction.html

_________________________________________

Correction: -ln²2/2 cannot be simplified to -ln2. Consequently, instead of the above, the sum will be

-ln²2/2 + ln 2·γ = 1/2·(2γ – ln 2)·ln 2, exactly as Mathematica said.

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