Question Number 211594 by mnjuly1970 last updated on 14/Sep/24
$$ \\ $$$$\:{If}\:,\:\:\overset{\:\:−} {{H}}_{{n}} \:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:−…+\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\:\:\: \\ $$$${prove}\:{that}:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\overset{\:\:−\:} {{H}}_{{n}} −\mathrm{ln}\left(\mathrm{2}\right)}{{n}}=\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:−−−−−−−−−−\:\:\:\:\:\: \\ $$
Commented by Ghisom last updated on 14/Sep/24
![just a hint: H_n ^ =Σ_(j=1) ^n (((−1)^(j+1) )/j) ⇒ lim_(n→∞) H_n ^ =ln 2 [Taylor series of ln (x+1); ∣x∣<1∨x=1]](https://www.tinkutara.com/question/Q211604.png)
$$\mathrm{just}\:\mathrm{a}\:\mathrm{hint}: \\ $$$$\bar {{H}}_{{n}} =\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{j}+\mathrm{1}} }{{j}}\:\Rightarrow\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\bar {{H}}_{{n}} \:=\mathrm{ln}\:\mathrm{2} \\ $$$$\left[\mathrm{Taylor}\:\mathrm{series}\:\mathrm{of}\:\mathrm{ln}\:\left({x}+\mathrm{1}\right);\:\mid{x}\mid<\mathrm{1}\vee{x}=\mathrm{1}\right] \\ $$