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Question Number 107609 by mathdave last updated on 11/Aug/20

Answered by mr W last updated on 11/Aug/20

$$\underset{k=1}{\sum ^n}\frac{1}{k+4}$$$$=\underset{k=5}{\sum ^{n+4}}\frac{1}{k}$$$$=\underset{k=1}{\sum ^{n+4}}\frac{1}{k}-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})$$$$=H_{n+4}-\frac{25}{12}$$

Commented by mathdave last updated on 11/Aug/20

$$ireallyappreciateddsthankx$$

Answered by hgrocks last updated on 11/Aug/20

$${\mathrm{S}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}+\mathrm{x}}=\underset{\mathrm{k}=1+\mathrm{x}}{\sum ^{\mathrm{n}+\mathrm{x}}}\frac{1}{\mathrm{k}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}+\mathrm{x}}}\frac{1}{\mathrm{k}}-\underset{\mathrm{k}=1}{\sum ^{\mathrm{x}}}\frac{1}{\mathrm{k}}$$$${\mathrm{S}}_{\mathrm{n}}={\mathrm{H}}_{\mathrm{n}+\mathrm{x}}-{\mathrm{H}}_{\mathrm{x}}$$$$$$$$$$

Commented by mathdave last updated on 11/Aug/20

$$thanksosomuch$$

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