1. Statement of the problem. We derive Eq. (13.42), namely $$\phi^{(\lambda)}(k,H)=\epsilon_k^{\lambda-1}\sum_{n=0}^{\infty}[-\tau(\epsilon_k)]^{n+1}\omega_c^{n}\Omega_k^{n}v_k.$$ 2. Start from the substituted equation obtained by inserting the second of (13.39) into the first. The resulting equation for $g^{(\lambda)}$ reads $$g^{(\lambda)}=-\epsilon_k^{\lambda-1}\tau(\epsilon_k)v_k+\tau^2(\epsilon_k)\omega_c^2\Omega_k^2\,g^{(\lambda)}.$$ 3. Expand $g^{(\lambda)}$ as a power series in $\omega_c^2$ and identify the zeroth term. Write $$g^{(\lambda)}=\sum_{l=0}^{\infty}g_l^{(\lambda)}\omega_c^{2l},\quad g_0^{(\lambda)}=-\epsilon_k^{\lambda-1}\tau(\epsilon_k)v_k.$$ 4. Insert the series into the equation and equate coefficients of like powers of $\omega_c^2$. This yields the recurrence for $l\ge 1$: $$g_l^{(\lambda)}=\tau^2(\epsilon_k)\Omega_k^2\,g_{l-1}^{(\lambda)}.$$ 5. Solve the recurrence by iteration and substitute $g_0^{(\lambda)}$. By iteration we get $$g_l^{(\lambda)}=\tau^{2l}(\epsilon_k)\Omega_k^{2l}g_0^{(\lambda)},$$ so substituting $g_0^{(\lambda)}$ gives $$g^{(\lambda)}=\epsilon_k^{\lambda-1}\sum_{l=0}^{\infty}[-\tau(\epsilon_k)]^{2l+1}\omega_c^{2l}\Omega_k^{2l}v_k.$$ 6. Obtain the analogous expansion for $u^{(\lambda)}$ from the second equation in (13.39). One finds $$u^{(\lambda)}=\epsilon_k^{\lambda-1}\sum_{l=0}^{\infty}[-\tau(\epsilon_k)]^{2l+2}\omega_c^{2l+1}\Omega_k^{2l+1}v_k.$$ 7. Combine $g^{(\lambda)}$ and $u^{(\lambda)}$ into a single summation over $n$. Take even $n=2l$ from $g^{(\lambda)}$ and odd $n=2l+1$ from $u^{(\lambda)}$ and merge both series to obtain a single sum over $n\ge 0$. This yields $$\phi^{(\lambda)}(k,H)=g^{(\lambda)}+u^{(\lambda)}=\epsilon_k^{\lambda-1}\sum_{n=0}^{\infty}[-\tau(\epsilon_k)]^{n+1}\omega_c^{n}\Omega_k^{n}v_k.$$ 8. Final answer. $$\boxed{\phi^{(\lambda)}(k,H)=\epsilon_k^{\lambda-1}\sum_{n=0}^{\infty}[-\tau(\epsilon_k)]^{n+1}\omega_c^{n}\Omega_k^{n}v_k.}$$