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Question Number 201860 by MrGHK last updated on 14/Dec/23
𝚺_(n=1) ^∞ (((−1)^n H_n )/(n+1))=??
$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}+\mathrm{1}}=?? \\ $$
Answered by mnjuly1970 last updated on 14/Dec/23
    Σ_(n=1) ^∞ H_n x^n  = −((ln(1−x))/(1−x))       where ,  H_n = 1+(1/2)+(1/3) +...+(1/2)          ∫_0 ^( x) Σ_(n=1) ^∞ H_n t^( n) dt = (1/2) ln^2 (1−x) + C              Σ_(n=1) ^∞  ((H_n x^(n+1) )/(n+1)) = (1/2) ln^2 (1−x) +C    x=0 ⇒  C=0         Σ_(n=) ^∞ (( H_n x^(n+1) )/(n+1)) = (1/2)ln^2 (1−x)        x=−1⇒ Σ_(n=1) ^∞ (( (−1)^n H_n )/(n+1)) =− (1/2) ln^2 (2)
$$\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {x}^{{n}} \:=\:−\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}} \\ $$$$\:\:\:\:\:{where}\:,\:\:{H}_{{n}} =\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\:+…+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:{x}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {t}^{\:{n}} {dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\:+\:{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{H}_{{n}} {x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\:+{C} \\ $$$$\:\:{x}=\mathrm{0}\:\Rightarrow\:\:{C}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\underset{{n}=} {\overset{\infty} {\sum}}\frac{\:{H}_{{n}} {x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\: \\ $$$$\:\:\:\:\:{x}=−\mathrm{1}\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}+\mathrm{1}}\:=−\:\frac{\mathrm{1}}{\mathrm{2}}\:{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$ \\ $$
Commented by MrGHK last updated on 14/Dec/23
 so we can say 𝚺_(n=1) ^∞ ((H_n x^(n+1) )/(n+1))=(1/2)ln^2 (1−x)
$$\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{say}}\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\boldsymbol{\mathrm{n}}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right) \\ $$

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