Question

Example 16 Sum to infinity of the series \frac{2\cdot3}{3!} + \frac{3\cdot5}{4!} + \frac{4\cdot7}{5!} + \frac{5\cdot9}{6!} + ... Solution: To find nth term, the first factor and second factor in each term form an arith progression. (Numerator and denominator also) Then the nth term becomes $t_n = \frac{(n+1)(2n+1)}{(n+2)!}$ Let $(n+1)(2n+1) = A(n+1)(n+2) + B(n+2) + C$ Equating like coefficients we obtain A = 2, B = -3, C = 3 Then $t_n = \frac{2}{n!} - \frac{3}{(n+1)!} + \frac{3}{(n+2)!}$ Putting n = 1, 2, 3, ... $t_1 = \frac{2}{1!} - \frac{3}{2!} + \frac{3}{3!}$ $t_2 = \frac{2}{2!} - \frac{3}{3!} + \frac{3}{4!}$ $t_3 = \frac{2}{3!} - \frac{3}{4!} + \frac{3}{5!}$

          Example 16
Sum to infinity of the series \frac{2\cdot3}{3!} + \frac{3\cdot5}{4!} + \frac{4\cdot7}{5!} + \frac{5\cdot9}{6!} + ...
Solution: To find nth term, the first factor and second factor in each term form an arith
progression. (Numerator and denominator also)
Then the nth term becomes $t_n = \frac{(n+1)(2n+1)}{(n+2)!}$
Let $(n+1)(2n+1) = A(n+1)(n+2) + B(n+2) + C$
Equating like coefficients we obtain
A = 2, B = -3, C = 3
Then
$t_n = \frac{2}{n!} - \frac{3}{(n+1)!} + \frac{3}{(n+2)!}$
Putting n = 1, 2, 3, ...
$t_1 = \frac{2}{1!} - \frac{3}{2!} + \frac{3}{3!}$
$t_2 = \frac{2}{2!} - \frac{3}{3!} + \frac{3}{4!}$
$t_3 = \frac{2}{3!} - \frac{3}{4!} + \frac{3}{5!}$

Example 16
Sum to infinity of the series (2·3)/(3!) + (3·5)/(4!) + (4·7)/(5!) + (5·9)/(6!) + ...
Solution: To find nth term, the first factor and second factor in each term form an arith
progression. (Numerator and denominator also)
Then the nth term becomes tn = ((n+1)(2n+1))/((n+2)!)
Let (n+1)(2n+1) = A(n+1)(n+2) + B(n+2) + C
Equating like coefficients we obtain
A = 2, B = -3, C = 3
Then
tn = (2)/(n!) - (3)/((n+1)!) + (3)/((n+2)!)
Putting n = 1, 2, 3, ...
t1 = (2)/(1!) - (3)/(2!) + (3)/(3!)
t2 = (2)/(2!) - (3)/(3!) + (3)/(4!)
t3 = (2)/(3!) - (3)/(4!) + (3)/(5!)

Added by Jose M.

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