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Let $\sum_{n=0}^\infty a_ne^{-\lambda_n x}$, where $0 < \lambda_n < \lambda_{n+1}$. It is given that the series converges at $x_0$. Prove that the series converges uniformly at $[x_0,\infty)$. Guidance: Prove it first when $x_0 = 0$.

So I followed the guidance:

$$\sum_{n=0}^\infty a_ne^{-\lambda_n}\cdot 0 = \sum_{n=0}^\infty a_ne^0 = \sum_{n=0}^\infty a_n < \infty$$

Now, how can I tell that for every $x>0$ the series converges uniformly? As far as I can understand the term $e^{-\lambda_nx}$ can be as big as we want.

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Assuming the $a_n$'s are non-negative (or that we have absolute convergence at $x_0$), then notice that $|a_n e^{-\lambda_n x }| \leq a_n e^{-\lambda x_0}$ for any $x \in [x_0,\infty)$. Since $\sum_{n=0}^{\infty} a_n e^{-\lambda_n x_0} < \infty$, the Weierstrass M-Test implies that $\sum_{n=0}^{\infty} a_n e^{-\lambda_n x}$ converges uniformly for $x \in [x_0,\infty)$.

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