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Test for convergence of series \[ \frac{1}{1 \cdot 2 \cdot 3}+\frac{3}{2 \cdot 3 \cdot 4}+\frac{5}{3 \cdot 4 \cdot 5}+\ldots . \infty \] |
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Answer» \(\frac1{1.2.3}+\frac3{2.3.4}+\frac5{3.4.5}+...+\frac{2n-1}{n(n+1)(n+2)}+....\infty\) \(\therefore\) nth term of series is Tn = \(\frac{2n-1}{n(n+1)(n+2)}\) = \(\frac{2n}{n(n+1)(n+2)}-\frac{1}{n(n+1)(n+2)}\) = \(\frac2{(n+1)(n+2)}-\frac1{n(n+1)(n+2)}\) = 2\(\left(\frac{(n+2)-(n+1)}{(n+1)(n+2)}\right)-\frac12\)\((\frac1n-\frac2{n+1}+\frac1{n+2})\) = 2\((\frac1{n+1}-\frac1{n+2})-\frac12((\frac 1n-\frac1{n+1})-(\frac1{n+1}-\frac1{n+2}))\) \(\therefore\) Sn = \(\sum\)Tn = 2\(\sum\)(\(\frac{1}{n+1}-\frac1{n+2}\)) - \(\frac12\)\(\sum\)(\(\frac1n-\frac1{n+1}\)) + \(\frac12\)\(\sum\)(\(\frac1{n+1}-\frac1{n+2}\)) = \(\frac52(\frac12-\frac13+\frac13-\frac14+....-\frac1{n+1}+\frac1{n+2}....\infty)\) - \(\frac12(1-\frac12+\frac12-\frac13+....-\frac1n+\frac 1n-\frac1{n+1}+.....\infty)\) = \(\frac52\times\frac12-\frac12\times1 = \frac54-\frac12=\frac{5-2}4=\frac34\) Hence, given series is convergent and convergent to to 3/4 |
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