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Question Number 162478 by HongKing last updated on 29/Dec/21

Calculate: Σ_(k=1) ^∞ ((H_k 2^(-k) )/(k + 1)) where H_k is the k-th harmonic number

$$\mathrm{Calculate}:\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{H}_{\boldsymbol{\mathrm{k}}} \:\mathrm{2}^{-\boldsymbol{\mathrm{k}}} }{\mathrm{k}\:+\:\mathrm{1}} \\ $$$$\mathrm{where}\:\mathrm{H}_{\boldsymbol{\mathrm{k}}} \:\mathrm{is}\:\mathrm{the}\:\boldsymbol{\mathrm{k}}-\mathrm{th}\:\mathrm{harmonic}\:\mathrm{number} \\ $$

Answered by mnjuly1970 last updated on 29/Dec/21

−−solution−− Ψ=Σ_(n=1) ^∞ H_n x^( n) = −((ln(1−x))/(1−x)) integration both sides ΣH_n ∫_0 ^( x) t^( n) dt = (1/(2 )) ln^( 2) (1−x) Σ_(n=1) ^∞ (( H_( n) x^( n+1) )/(n+1)) = (1/2) ln^( 2) (1−x) x= (1/2) ⇒ Σ_(n=1) ^∞ (( H_( n) )/(( n+1 ).2^( n) )) = ln^( 2) ((1/2) ) Ψ= ln^( 2) ( 2 )

$$\:\:−−{solution}−− \\ $$$$\:\:\:\:\:\:\Psi=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{H}_{{n}} {x}^{\:{n}} \:=\:−\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}} \\ $$$$\:\:\:\:\:\:\:{integration}\:{both}\:{sides} \\ $$$$\:\:\:\:\:\:\Sigma\mathrm{H}_{{n}} \int_{\mathrm{0}} ^{\:{x}} {t}^{\:{n}} {dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}\:}\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{H}_{\:{n}} {x}^{\:{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\right) \\ $$$$\:\:\:\:\:\:\:{x}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{H}_{\:{n}} }{\left(\:{n}+\mathrm{1}\:\right).\mathrm{2}^{\:{n}} }\:=\:{ln}^{\:\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Psi=\:{ln}^{\:\mathrm{2}} \left(\:\mathrm{2}\:\right)\:\:\: \\ $$

Commented by HongKing last updated on 31/Dec/21

Cool my dear Sir thank you so much

$$\mathrm{Cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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