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Question Number 84835 by Power last updated on 16/Mar/20

Commented by abdomathmax last updated on 16/Mar/20

$$S=\sum _{k=1}^{32}\frac{k}{k+1}=\sum _{k=1}^{32}\frac{k+1-1}{k+1}$$$$=\sum _{k=1}^{32}\left(1\right)-\sum _{k=1}^{32}\frac{1}{k+1}$$$$=32-\sum _{k=2}^{33}\frac{1}{k}=32-\sum _{k=1}^{33}\frac{1}{k}+1$$$$=34-H_{33}with{H}_n=\sum _{k=1}^n\frac{1}{k}$$

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