Question Number 104604 by ~blr237~ last updated on 22/Jul/20

Σ_(n=1) ^∞ (1+(1/2)+...+(1/n))(1/2^n ) = ln4

Answered by OlafThorendsen last updated on 22/Jul/20

S = Σ_(n=1) ^∞ (1+(1/2)+...+(1/n))(1/2^n ) S = Σ_(n=1) ^∞ (1/2^n )Σ_(k=1) ^n (1/k) S = Σ_(n=1) ^∞ (1/n)Σ_(k=n) ^∞ (1/2^k ) S = lim_(N→∞) Σ_(n=1) ^∞ (1/n).(1/2^n ).((1−((1/2))^N )/(1−(1/2))) S = 2Σ_(n=1) ^∞ (1/n).(1/2^n ) S = 2Σ_(n=1) ^∞ (2^(−n) /n) (1/(1−x)) = Σ_(n=0) ^∞ x^k , ∣x∣<1 −ln∣1−x∣ = Σ_(n=1) ^∞ (x^k /k) S = 2Σ_(n=1) ^∞ (2^(−n) /n) = −2ln∣1−(1/2)∣ S = 2Σ_(n=1) ^∞ (2^(−n) /n) = −2ln(1/2) = ln4

Commented by~blr237~ last updated on 23/Jul/20

Nice work Sir