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## Homework Statement

The Taylor expansion of ln(1+x) has terms which decay as 1/n.

Show, that by choosing an appropriate constant 'c', the Taylor series of

(1+cx)ln(1+x)

can be made to decay as 1/n

^{2}

(assuming expansion about x=0)

## Homework Equations

f(x)=[itex]\sum[/itex][itex]^{n=\infty}_{n=0}[/itex] f

^{(n)}(0) [itex]\frac{x^{n}}{n!}[/itex]

## The Attempt at a Solution

I used Maple to differentiate this function and find values at x=0 for several derivative:

f

^{(0)}(0) = 0

f

^{(1)}(0) = 1

f

^{(2)}(0) = 2c-1

f

^{(3)}(0) = -3c+2

f

^{(4)}(0) = 8c-6

f

^{(5)}(0) = -30c+24

f(x)=[itex]\frac{(0)x^{0}}{0!}[/itex]+[itex]\frac{(1)x^{1}}{1!}[/itex]+[itex]\frac{(2c-1)x^{2}}{2!}[/itex]+[itex]\frac{(-3c+2)x^{3}}{3!}[/itex]+[itex]\frac{(8c-6)x^{4}}{4!}[/itex]+[itex]\frac{(-30c+24)x^{5}}{5!}[/itex] ...

This is where I'm stuck... In order to get the terms decaying as 1/n

^{2}, I get different values of c for each term...

c0=1

c1=1

c2=[itex]\frac{3}{4}[/itex]

c3=[itex]\frac{8}{9}[/itex]

c4=[itex]\frac{15}{16}[/itex]

c5=[itex]\frac{24}{25}[/itex]

And I need one constant c that will do it all. Any help would be greatly appreciated.