# 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.

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Why is: $$\sum_{n=1}^M \frac{a_n}{e^{nt}} \leq \sum_{n=1}^M \frac{C}{e^{nt}}$$? –  Thomas Andrews Jan 3 '13 at 19:40
Well, for every $n\in N$ we have that $a_n < C$ and the series converges since the other converges –  V. Galerkin Jan 3 '13 at 19:43
Not necessarily, actually $|a_n|\leq 2C$. (You also need to be careful with your absolute values.) For example, if $a_0=C$ and $a_n=(-1)^n(2C)$ for $n>0$, the partial sums are bounded above by $C$. –  Thomas Andrews Jan 3 '13 at 19:44
Need to make more careful use of absolute values. Or else let $b_n=|a_n|$ and prove absolute convergence. Boundedness of the $|a_n|$ needs more detail. –  André Nicolas Jan 3 '13 at 19:44

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 $$\sum_{n=1}^N a(n) b(n) = \sum_{n=1}^N b(n)(A(n)-A(n-1)) = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\ = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1}^N b(n+1)A(n) = b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ 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.