Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 166254 by mnjuly1970 last updated on 16/Feb/22

      Θ=Σ_(n=1) ^∞ (( H_( n) )/(n. (n+1 )))  =^?  (π^( 2) /6)       −−−−+

$$ \\ $$$$\:\:\:\:\Theta=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{H}_{\:{n}} }{{n}.\:\left({n}+\mathrm{1}\:\right)}\:\:\overset{?} {=}\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{6}} \\ $$$$\:\:\:\:\:−−−−+ \\ $$

Answered by Kamel_Ben last updated on 16/Feb/22

Θ=Σ_(n=1) ^(+∞) ((H_(n+1) −(1/(n+1)))/(n(n+1)))=Σ_(n=1) ^(+∞) (H_(n+1) /(n(n+1)))−Σ_(n=1) ^(+∞) (1/(n(n+1)))+Σ_(n=1) ^(+∞) (1/((n+1)^2 ))     =Σ_(n=1) ^(+∞) ((H_n /n)−(H_(n+1) /(n+1)))+(π^2 /6)−1=(π^2 /6)

$$\Theta=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{{H}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}}{{n}\left({n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{{H}_{{n}+\mathrm{1}} }{{n}\left({n}+\mathrm{1}\right)}−\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}+\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{{H}_{{n}} }{{n}}−\frac{{H}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com