If the partial sums of a $a_n$ are bounded, then $\sum{}_{n=1}^\infty a_n e^{-nt}$ converges for all $t > 0$ 
If the partial sums of a $a_n$ are bounded, then $$\sum_{n=1}^\infty \frac{a_n }{e^{nt}}$$ converges for all $t > 0$.

Proof: since the partial sums of $a_n$ are bounded, then exists $C > 0 $ such that $\left|\sum_{n=1}^M a_n \right| < C$ forall $M \in \mathbb{N}$, so for every $n\in \mathbb N $ we have that $|a_n| < C$. Now,
$$ \sum_{n=1}^\infty \frac{a_n}{e^{nt}} = \lim_{M\to\infty} \sum_{n=1}^M \frac{a_n}{e^{nt}} \leq \lim_{M\to\infty} \sum_{n=1}^M \frac{C}{e^{nt}} = C \sum_{n=1}^\infty \left(\frac{1}{e^t}\right)^n < \infty \iff \frac{1}{e^t} < 1 $$
and this is satisfied for every $t > 0$. Is there any error? I'm not convinced with my way of writing the proof. Thanks in advance.
 A: You can conclude it based on Abel partial summation (The result is termed as generalized alternating test or Dirichlet test). We will prove the generalized statement first.
Consider the sum $S_N = \displaystyle \sum_{n=1}^N a(n)b(n)$. Let $A(n) = \displaystyle \sum_{n=1}^N a(n)$. If $b(n) \downarrow 0$ and $A(n)$ is bounded, then the series $\displaystyle \sum_{n=1}^{\infty} a(n)b(n)$ converges.
First note that from Abel summation, we have that
\begin{align*}\sum_{n=1}^N a(n) b(n) &= \sum_{n=1}^N b(n)(A(n)-A(n-1))\\&= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\
\\&= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1} b(n+1)A(n) \\&= b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))\end{align*}
Now if $A(n)$ is bounded i.e. $\vert A(n) \vert \leq M$ and $b(n)$ is decreasing, then we have that
$$\sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1)) \leq \sum_{n=1}^{N-1} M (b(n)-b(n+1))\\ = M (b(1) - b(N)) \leq Mb(1)$$
Hence, we have that $\displaystyle \sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1))$ converges and hence $$\displaystyle \sum_{n=1}^{N-1} A(n)  (b(n)-b(n+1))$$ converges absolutely. Now since
$$\sum_{n=1}^N a(n) b(n) = b(N) A(N) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$
we have that $\displaystyle \sum_{n=1}^N a(n)b(n)$ converges. In your case, $a(n) = a_n$. Hence, $A(N) = \displaystyle \sum_{n=1}^N a(n)$ is bounded. Also, $b(n) = \dfrac1{e^{nt}}$ is a monotone decreasing sequence converging to $0$ for $t>0$.
Hence, we have that $$\sum_{n=1}^N a_n e^{-nt}$$ converges.
