Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.