Question Number 112217 by mathdave last updated on 06/Sep/20
![proporsed by m.n july 1970 evaluate Σ_(n=1) ^∞ (H_n /(n2^n )) solution recall thatΣ_(n=1) ^∞ H_n x^n =((ln(1−x))/(x−1)) ∵((H_n x^n )/n)=∫_0 ^x H_n t^(n−1) dt Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=∫_0 ^x ((ln(1−t))/(t(t−1)))dt Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=∫_0 ^x ((ln(1−t))/(t−1))dt−∫_0 ^x ((ln(1−t))/t)dt=A−B let A=∫_0 ^x ((ln(1−t))/(t−1))dt=−∫_0 ^x ((ln(1−t))/(1−t))dt A=−[−ln^2 (1−t)]_0 ^x +∫_0 ^x −((ln(1−t))/(1−t))dt 2A=ln^2 (1−x) A=(1/2)ln^2 (1−x)......(1) then B=∫_0 ^x ((ln(1−t))/t)dt=∫_0 ^1 ((ln(1−t))/t)dt+∫_1 ^x ((ln(1−t))/t)dt B=−Li_2 (1)−[Li_2 (x)−Li_2 (1)] B=−Li_2 (x)....(2) but I=A−B I=Σ_(n=1) ^∞ ∫_0 ^x H_n t^(n−1) dt=(1/2)ln^2 (2)+Li_2 (x) ∵Σ_(n=1) ^∞ ((H_n x^n )/n)=(1/2)ln^2 (2)+Li_2 (x) but x=(1/2) Σ_(n=1) ^∞ (H_n /(n2^n ))=(1/2)ln^2 (2)+Li_2 ((1/2))=(1/2)ln^2 (2)+(π^2 /(12))−(1/2)ln^2 (2) ∵Σ_(n=1) ^∞ (H_n /(n2^n ))=(π^2 /(12)) by mathdave(06/09/2020)](Q112217.png)
$${proporsed}\:{by}\:{m}.{n}\:{july}\:\mathrm{1970} \\ $$$${evaluate} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}\mathrm{2}^{{n}} } \\ $$$${solution}\: \\ $$$${recall}\:{that}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {x}^{{n}} =\frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}−\mathrm{1}} \\ $$$$\because\frac{{H}_{{n}} {x}^{{n}} }{{n}}=\int_{\mathrm{0}} ^{{x}} {H}_{{n}} {t}^{{n}−\mathrm{1}} {dt} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{{x}} {H}_{{n}} {t}^{{n}−\mathrm{1}} {dt}=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{{x}} {H}_{{n}} {t}^{{n}−\mathrm{1}} {dt}=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}−\mathrm{1}}{dt}−\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}={A}−{B} \\ $$$${let}\:{A}=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}−\mathrm{1}}{dt}=−\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}−{t}}{dt} \\ $$$${A}=−\left[−\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)\right]_{\mathrm{0}} ^{{x}} +\int_{\mathrm{0}} ^{{x}} −\frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}−{t}}{dt} \\ $$$$\mathrm{2}{A}=\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right) \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)......\left(\mathrm{1}\right) \\ $$$${then}\: \\ $$$${B}=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}+\int_{\mathrm{1}} ^{{x}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$$${B}=−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)−\left[{Li}_{\mathrm{2}} \left({x}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)\right] \\ $$$${B}=−{Li}_{\mathrm{2}} \left({x}\right)....\left(\mathrm{2}\right) \\ $$$${but}\:{I}={A}−{B} \\ $$$${I}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{{x}} {H}_{{n}} {t}^{{n}−\mathrm{1}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{Li}_{\mathrm{2}} \left({x}\right) \\ $$$$\because\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} {x}^{{n}} }{{n}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{Li}_{\mathrm{2}} \left({x}\right) \\ $$$${but}\:\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}\mathrm{2}^{{n}} }=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\because\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}\mathrm{2}^{{n}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${by}\:{mathdave}\left(\mathrm{06}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 06/Sep/20
$${thank}\:{you}\:{so}\:{much}\:\:{mr}\: \\ $$$${mathdave}\:\:...{grateful}. \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$