Question Number 161822 by HongKing last updated on 22/Dec/21
![Find: Ω =lim_(n→∞) Σ_(k=1) ^n [Σ_(i=1) ^k (i + (1/4))]^( -1) = ?](Q161822.png)
$$\mathrm{Find}: \\ $$$$\Omega\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\left[\underset{\boldsymbol{\mathrm{i}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{k}}} {\sum}}\left(\mathrm{i}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\right]^{\:-\mathrm{1}} =\:? \\ $$
Answered by qaz last updated on 23/Dec/21
![Σ_(k=1) ^∞ [Σ_(i=1) ^k (i+(1/4))]^(−1) =Σ_(k=1) ^∞ [((k(k+1))/2)+(k/4)]^(−1) =(8/3)Σ_(k=1) ^∞ ((1/(2k))−(1/(2k+3))) =(4/3)H_(3/2) =((32)/9)−(8/3)ln2](Q161841.png)
$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{k}} {\sum}}\left(\mathrm{i}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right]^{−\mathrm{1}} \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}}+\frac{\mathrm{k}}{\mathrm{4}}\right]^{−\mathrm{1}} \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2k}}−\frac{\mathrm{1}}{\mathrm{2k}+\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\mathrm{H}_{\mathrm{3}/\mathrm{2}} \\ $$$$=\frac{\mathrm{32}}{\mathrm{9}}−\frac{\mathrm{8}}{\mathrm{3}}\mathrm{ln2} \\ $$
Commented by HongKing last updated on 25/Dec/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$