Let $s = \sigma+it$ Let $s_0$ be such that $f \left({s_0}\right)$ converges.  Let $S\left({m, n}\right) = \displaystyle \sum_{ k\mathop = n}^m a_k e^{-\lambda_k s_0}$ We may create a new Dirichlet series that converges at 0 by writing: {{begin-eqn}} {{eqn | l = g \left({s}\right)       | r = f \left({s+s_0}\right)   }} {{eqn | r = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n \left({s + s_0}\right)} }} {{eqn | r = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_ns_0} e^{-\lambda_ns} }} {{end-eqn}} Thus it suffices to show $g \left({s}\right)$ converges uniformly for for $\left\vert \arg\left(s\right)\right\vert \le  a < \frac \pi 2$ By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon>0$ there exists an $N$ independent of $s$ such that for all $m,n>N$:  :$ \left\vert \displaystyle \sum_{k \mathop = n}^m a_n e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert < \epsilon$ By Abel's Lemma: Formulation 2 we may write: {{begin-eqn}} {{eqn | l = \left\vert \sum_{k \mathop = n}^m a_k e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert       | r = \left\vert \sum_{k \mathop = n}^m \left(S\left(k,n\right) - S\left(k-1,n\right)\right) e^{-\lambda_ks} \right\vert }} {{eqn | r = \left\vert S\left(m,n\right)e^{-\lambda_ms} + \sum_{k \mathop = n}^{m-1} S\left(k,n\right) \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert }} {{eqn | o = \le       | r = \left\vert S\left(m,n\right) e^{-\lambda_ms} \right\vert + \sum_{k \mathop = n}^{m-1} \left\vert S\left(k,n\right) \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert        | c = Triangle Inequality }} {{end-eqn}} Because $S\left(k,j\right)$ is the difference of two terms of a  convergent, and thus cauchy, sequence, we may pick $N$ large enough so that for $j>N$ :$\displaystyle \left\vert S\left(k,j\right) \right\vert < \frac {\epsilon \cos\left(a\right)}{3}$ which gives us: {{begin-eqn}} {{eqn | l = \left\vert S\left(m,n\right) e^{-\lambda_ms} \right\vert + \sum_{k \mathop = n}^{m-1} \left\vert S\left(k,n\right) \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert        | o = \le       | r = \frac {\epsilon \cos\left(a\right)}{3} \left(\left\vert e^{-\lambda_ms} \right\vert +  \sum_{k \mathop = n}^{m-1} \left\vert \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert  \right) }} {{end-eqn}} We see that: {{begin-eqn}} {{eqn | l = \left\vert e^{-\lambda_ks} - e^{-\lambda_{k+1}s} \right\vert       | r = \left\vert \int_{\lambda_k}^{\lambda_{k+1} } -se^{-xs} \rd x \right\vert }} {{eqn | o = \leq       | r = \int_{\lambda_k}^{\lambda_{k+1} } \left\vert-se^{-xs}\right\vert \rd x       | c = Modulus of Complex Integral }} {{eqn | r = \int_{\lambda_k}^{\lambda_{k+1} } \left\vert s \right\vert e^{-x\sigma} \rd x       | c =  }} {{eqn | r = \left\vert s \right\vert \int_{\lambda_k}^{\lambda_{k+1} } e^{-x\sigma} \rd x       | c =  }} {{eqn | r = \frac {\left\vert s \right\vert} {\sigma} \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)       | c =  }} {{eqn | r = \sec\left(\arg\left(s\right)\right) \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)       | c =  }} {{end-eqn}} From Shape of Secant Function, we have that on the interval $\left( -\frac \pi 2 , \frac \pi 2\right)$: :$ \left\vert \arg\left(s\right) \right\vert \le a \implies \sec\left(\arg\left(s\right)\right) \le \sec\left(a\right)$ which gives us: :$ \sec\left(\arg\left(s\right)\right) \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right) \le \sec\left(a\right) \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)$ Hence: {{begin-eqn}} {{eqn | l = \frac {\epsilon \cos\left(a\right)}{3} \left(\left\vert e^{-\lambda_ms} \right\vert +  \sum_{k \mathop = n}^{m-1} \left\vert \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert  \right)       | o = \le       | r = \frac {\epsilon \cos\left(a\right)}{3} \left( e^{-\lambda_m\sigma} + \sec\left(a\right) \sum_{k \mathop = n}^{m-1}  e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right) }} {{eqn | r = \frac {\epsilon \cos\left(a\right)}{3} \left( e^{-\lambda_m\sigma} + \sec\left(a\right) \left( e^{-\lambda_n\sigma} - e^{-\lambda_{m}\sigma}  \right) \right)       | c =  Telescoping Sum }} {{end-eqn}} Because $\sigma>0$, we have that $ - \lambda_k \sigma <0$ and hence: :$ e^{-\lambda_k \sigma} < 1 \le \sec\left(a\right) $ which gives us: {{begin-eqn}} {{eqn | l = \left\vert \displaystyle \sum_{k \mathop = n}^m a_n e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert       | o = \le       | r = \frac {\epsilon \cos\left(a\right)}{3} \left( e^{-\lambda_m\sigma} + \sec\left(a\right) \left( e^{-\lambda_n\sigma} - e^{-\lambda_{m}\sigma}  \right) \right) }} {{eqn | o = \le       | r = \frac {\epsilon \cos\left(a\right)}{3} \left( \sec\left(a\right) + \sec\left(a\right) \left( 1 + 1  \right) \right) }} {{eqn | r = \frac {\epsilon \cos\left(a\right)}{3} \left(  3 \sec\left(a\right) \right) }} {{eqn | r = \epsilon }} {{end-eqn}} {{qed}}