Steps in Proof of Theorem 11.3

$$\begin{gathered} |h(N)|=|-N^{s_k}||\sum _{n=N+1}^{\mathrm{\infty }}h(n)n^{-s_k}| \\ =N^{{\sigma }_k}|\sum _{n=N+1}^{\mathrm{\infty }}h(n)n^{-s_k}|. \end{gathered}$$

Substituting $\sigma ={\sigma }_k$ and $N=N+1$ in Lemma 1, we get

$$\begin{array}{rl}|h(N)|& \le N^{{\sigma }_k}(N+1{)}^{-({\sigma }_k-c)}\sum _{n=N+1}^{\mathrm{\infty }}|h(n)|n^{-c}\\ & =\frac{N^{{\sigma }_k}}{(N+1{)}^{{\sigma }_k}}\underset{A}{\underbrace{(N+1{)}^c\underset{\text{converges}}{\underbrace{\sum _{n=N+1}^{\mathrm{\infty }}|h(n)|n^{-c}}}}}\\ & ={\left(\frac{N}{N+1}\right)}^{{\sigma }_k}A.\end{array}$$