- #1

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## Homework Statement

Show that the first non-zero coefficient in the expansion of

##e^{-x}-\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}##

in ascending powers of x is that of x^5

## Homework Equations

Series expansion, logarithmic series

## The Attempt at a Solution

Let ##y=e^{-x}-\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}##

##{\ln y={\ln \left[\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}\right]\ }\ }##

##{\ln y\ }={\ln \left(1-x\right)\ }-\frac{1}{2}{\ln \left(1-x^2\right)-\frac{1}{3}{\ln \left(1-x^3\right)\ }\ }##

##{\ln y\ }=\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}\right)-\frac{1}{2}\left(-x^2-\frac{x^4}{2}\right)-\frac{1}{3}\left(-x^3\right)##

##{\ln y\ }=-x-\frac{x^5}{5}##

##y=e^{-x}\cdot e^{-\frac{x^5}{5}}##

##ye^x=e^{-\frac{x^5}{5}}##

##\left[\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}\right]e^x=e^{-\frac{x^5}{5}}##

I know I'm incredibly close to the answer, however it seems like I can't solve it.