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