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 $$\sum_{n=1}^{\infty}\frac{a_n}{e^n}$$ is convergent

denote$$S_n=\sum_{k=1}^{n}a_n$$

show that $$\sum_{n=1}^{\infty}\frac{S_n}{e^n}$$ is convergent.

share|improve this question
    
Just to clarify, $e=\ln(1)$ here, right? Its a constant? –  icurays1 Nov 14 '12 at 16:43
    
@icurays1 Rather $\ln e=1$. –  Hagen von Eitzen Nov 14 '12 at 16:49
    
Ha, whoops. Coffee hasn't kicked in yet. Yes, that's what I meant. –  icurays1 Nov 14 '12 at 16:51
    
yes, $e$ means euler constant –  Laura Nov 15 '12 at 0:45

1 Answer 1

up vote 3 down vote accepted

Use Abel's identity to show that:

$$\sum_{n=1}^{k}\frac{S_n}{e^n}=1/(e^{-1}-1)[e^{-(k+1)}S_{k+1}-e^{-1}S_1-\sum_{n=1}^{k}\frac{a_{n+1}}{e^{n+1}}]$$

Now the answer to your question follows very easily because the limit of the right hand side exists.

share|improve this answer
4  
+1 for nice argument! I want to comment that though the convergence of $S_{k} e^{-k}$ follows from Cesaro-Stolz theorem together with the observation $a_n = o(e^n)$, it still seems to deserve a justification when it comes to elementary level analysis. –  sos440 Nov 14 '12 at 16:53

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.