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Question Number 6487 by Temp last updated on 29/Jun/16

Prove and show why:  Σ_(k=1) ^n H_k =(n+1)(H_(n+1) −1)  Where:  H_k =1+(1/2)+(1/3)+...+(1/k)

$$\mathrm{Prove}\:\mathrm{and}\:\mathrm{show}\:\mathrm{why}: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{H}_{{k}} =\left({n}+\mathrm{1}\right)\left({H}_{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$\mathrm{Where}: \\ $$$${H}_{{k}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...+\frac{\mathrm{1}}{{k}} \\ $$

Commented by prakash jain last updated on 29/Jun/16

Σ_(k=1) ^n H_k =Σ_(k=1) ^n  Σ_(i=1) ^k (1/i) = Σ_(i=1) ^n ((n−i+1)/i)=Σ_(i=1) ^n (((n+1)/i)−1)  =(n+1)Σ_(i=1) ^n (1/i)−n=(n+1)(H_n )−n  =(n+1)H_n +1−n−1  =(n+1)(H_n +(1/(n+1)))−(n+1)  =(n+1)(H_(n+1) −1)

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{H}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{i}}\:=\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}−{i}+\mathrm{1}}{{i}}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{n}+\mathrm{1}}{{i}}−\mathrm{1}\right) \\ $$$$=\left({n}+\mathrm{1}\right)\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{i}}−{n}=\left({n}+\mathrm{1}\right)\left({H}_{{n}} \right)−{n} \\ $$$$=\left({n}+\mathrm{1}\right){H}_{{n}} +\mathrm{1}−{n}−\mathrm{1} \\ $$$$=\left({n}+\mathrm{1}\right)\left({H}_{{n}} +\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\left({n}+\mathrm{1}\right) \\ $$$$=\left({n}+\mathrm{1}\right)\left({H}_{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$

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