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Question Number 84109 by Power last updated on 09/Mar/20

Commented by mahdi last updated on 09/Mar/20

$$\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2\mathrm{x}+1}\approx \ln \left(\frac{2\mathrm{x}+1}{\sqrt{\mathrm{x}}}\right)+0.28861$$

Commented by mathmax by abdo last updated on 09/Mar/20

$$A_n=1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{n-2}$$$$case1n=2p+1\Rightarrow A_n=1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2p-1}$$$$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2p-1}+\frac{1}{2p}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{2p}$$$$=H_{2p}-\frac{1}{2}{H}_{p}wehavep=\left[\frac{n-1}{2}\right]\Rightarrow$$$$A_n=H_{2\left[\frac{n-1}{2}\right]}-\frac{1}{2}{H}_{\left[\frac{n-1}{2}\right]}$$$$case2n=2p\Rightarrow A_n=1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2(p-1)}$$$$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2p-4}+\frac{1}{2p-3}+\frac{1}{2(p-1)}$$$$-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{2(p-2)}=H_{2p-2}-\frac{1}{2}{H}_{p-2}$$$$=H_{2\left[\frac{n}{2}\right]-2}-\frac{1}{2}{H}_{\left[\frac{n}{2}\right]-2}$$$$H_N=\sum _{k=1}^N\frac{1}{k}$$

Answered by mind is power last updated on 09/Mar/20

$$\underset{k=0}{\sum ^n}\frac{1}{2k+1}=\underset{k=1}{\sum ^{2n+1}}\frac{1}{k}-\underset{k=1}{\sum ^n}\frac{1}{2k}=H_{2n+1}-\frac{H_n}{2}$$

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