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Question Number 103623 by Lordose last updated on 16/Jul/20

$$\mathbf{find}\mathbf{the}\mathbf{sum}\mathbf{of}\mathbf{the}\mathbf{series}\mathbf{whose}\mathbf{nth}$$$$\mathbf{term}\mathbf{is}\frac{2\mathbf{n}-1}{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2}.$$$$\mathbf{i}\mathbf{have}\mathbf{a}\mathbf{problem}\mathbf{with}\mathbf{this}\mathbf{and}\mathbf{i}\mathbf{need}$$$$\mathbf{help}\mathbf{please}$$

Answered by OlafThorendsen last updated on 16/Jul/20

$$\mathbf{find}\mathbf{the}\mathbf{sum}\mathbf{of}\mathbf{the}\mathbf{series}\mathbf{whose}\mathbf{nth}$$$$\mathbf{term}\mathbf{is}\frac{2\mathbf{n}-1}{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2}.$$$$\mathbf{i}\mathbf{have}\mathbf{a}\mathbf{problem}\mathbf{with}\mathbf{this}\mathbf{and}\mathbf{i}\mathbf{need}$$$$\mathbf{help}\mathbf{please}$$$$u_n=\frac{2n-1}{n(n+1)(n+2)}$$$$u_n=-\frac{1}{2}.\frac{1}{n}+\frac{3}{n+1}-\frac{5}{2}.\frac{1}{n+2}$$$$\underset{n=1}{\sum ^p}{u}_n=-\frac{1}{2}\underset{n=1}{\sum ^p}\frac{1}{n}+3\underset{n=1}{\sum ^p}\frac{1}{n+1}-\frac{5}{2}\underset{n=1}{\sum ^p}\frac{1}{n+2}$$$$\underset{n=1}{\sum ^p}{u}_n=-\frac{1}{2}\underset{n=1}{\sum ^p}\frac{1}{n}+3\underset{n=2}{\sum ^{p+1}}\frac{1}{n}-\frac{5}{2}\underset{n=3}{\sum ^{p+2}}\frac{1}{n}$$$$\underset{n=1}{\sum ^p}{u}_n=-\frac{1}{2}(1+\frac{1}{2})+3\left(\frac{1}{2}\right)+3\left(\frac{1}{p+1}\right)-\frac{5}{2}(\frac{1}{p+1}+\frac{1}{p+2})$$$$\underset{n=1}{\sum ^p}{u}_n=\frac{3}{4}+\frac{1}{2}\left(\frac{1}{p+1}\right)-\frac{5}{2}\left(\frac{1}{p+2}\right)$$$$\underset{n=1}{\sum ^p}{u}_n=\frac{3}{4}-\frac{2p+\frac{3}{2}}{(p+1)(p+2)}$$$$\mathrm{and}\underset{n=1}{\sum ^{\mathrm{\infty }}}{u}_n=\frac{3}{4}$$$$$$

Commented by Lordose last updated on 16/Jul/20

$$\mathbf{sir}\mathbf{i}{\mathbf{don}}^{\prime }\mathbf{t}\mathbf{understand}\mathbf{the}4\mathbf{th}\mathbf{line}$$$$\mathbf{can}\mathbf{you}\mathbf{explain}\mathbf{pls}$$

Answered by bobhans last updated on 16/Jul/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n-1}{n(n+1)(n+2)}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{3}{n+1}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2n}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{5}{2(n+2)}$$$$=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{3}{n}-\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{5}{2}\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{1}{n}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{3}{n}-3-\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{5}{2}\{\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-1-\frac{1}{2}\}$$$$=-3+\frac{15}{4}+(3-\frac{1}{2}-\frac{5}{2})\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}$$$$=\frac{3}{4}+0\underset{k\to \mathrm{\infty }}{\lim }\underset{n=1}{\sum ^k}\frac{1}{n}=\frac{3}{4}+0=\frac{3}{4}$$

Commented by Lordose last updated on 16/Jul/20

$$\mathbf{can}\mathbf{anyone}\mathbf{make}\mathbf{me}\mathbf{understand}\mathbf{pls}$$

Commented by bobhans last updated on 16/Jul/20

$$whatpartyou{don}^{\prime }tunderstand?$$

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