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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
up vote 5 down vote accepted

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.

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I have always found the Dirichlet Test intriguing. Every time I had to use it to solve a problem, I could find no other means of solution. – Haskell Curry Jan 3 '13 at 19:50
@HaskellCurry Yes. It is one of those intriguing and powerful mathematical tool. – user17762 Jan 3 '13 at 19:51

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