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Question-107609




Question Number 107609 by mathdave last updated on 11/Aug/20
Answered by mr W last updated on 11/Aug/20
Σ_(k=1) ^n (1/(k+4))  =Σ_(k=5) ^(n+4) (1/k)  =Σ_(k=1) ^(n+4) (1/k)−(1+(1/2)+(1/3)+(1/4))  =H_(n+4) −((25)/(12))
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{4}} \\ $$$$=\underset{{k}=\mathrm{5}} {\overset{{n}+\mathrm{4}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{4}} {\sum}}\frac{\mathrm{1}}{{k}}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$={H}_{{n}+\mathrm{4}} −\frac{\mathrm{25}}{\mathrm{12}} \\ $$
Commented by mathdave last updated on 11/Aug/20
i really appreciated ds  thankx
$${i}\:{really}\:{appreciated}\:{ds}\:\:{thankx} \\ $$
Answered by hgrocks last updated on 11/Aug/20
S_n = Σ_(k=1) ^n (1/(k+x)) =  Σ_(k=1+x) ^(n+x) (1/k) = Σ_(k=1) ^(n+x) (1/k) − Σ_(k=1) ^x (1/k)   S_n     = H_(n+x)  − H_x
$$\mathrm{S}_{\mathrm{n}} =\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}+\mathrm{x}}\:=\:\:\underset{\mathrm{k}=\mathrm{1}+\mathrm{x}} {\overset{\mathrm{n}+\mathrm{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}\:=\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}+\mathrm{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}\:−\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{x}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\:\mathrm{S}_{\mathrm{n}} \:\:\:\:=\:\mathrm{H}_{\mathrm{n}+\mathrm{x}} \:−\:\mathrm{H}_{\mathrm{x}} \\ $$$$ \\ $$$$ \\ $$
Commented by mathdave last updated on 11/Aug/20
thank so so much
$${thank}\:{so}\:{so}\:{much} \\ $$

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