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.
Best Answer
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.