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.

Trying to determine if this alternating series converges absolutely or conditionally. ATS criteria has been met (terms are positive [ignoring signs] & decreasing, and the lim n->inf = 0, assuming I haven't made a mistake) so I know it's at least convergent but need to prove absolute convergence. However, I believe that if the absolute value of the series is convergent then it is impossible for the original series to be divergent and, thus, the series has to be absolutely convergent. Does that make sense? Thanks for taking a look at this.

enter image description here

share|improve this question

2 Answers 2

up vote 2 down vote accepted

The terms, at least after a while, are decreasing in absolute value. However, there is no need to show that. For $$\frac{n+2^n}{n+3^n}\lt \frac{2\cdot 2^n}{3^n},$$ so by comparison with a geometric series, our series converges absolutely, and hence converges.

share|improve this answer
    
This is great but am I correct in saying that if the absolute value of an alternating series is convergent then the original alternating series is also convergent? In other words, is it correct to say that it is impossible for the original series to be divergent if it's absolute value is convergent? –  joe schmoe Jul 13 at 22:02
1  
Yes, you are correct. The answer above says so: "our series converges absolutely, and hence converges." The signs need not actually alternate. For any sequence $(a_n)$, if $\sum|a_n|$ converges, then $\sum a_n$ converges. Of course, the converse need not hold. If $\sum a_n$ converges, then $\sum|a_n|$ need not converge. –  André Nicolas Jul 13 at 22:06
    
You rock. Thank you! –  joe schmoe Jul 13 at 22:23
    
You are welcome. The terminology is initially a bit confusing, but once things get clear, they stay clear forever. –  André Nicolas Jul 13 at 22:47

Hint: $$\left|(-1)^{n}\frac{n+2^{n}}{n+3^{n}}\right|=\frac{1+\frac{n}{2^{n}}}{1+\frac{n}{3^{n}}}\left(\frac{2}{3}\right)^{n}$$

share|improve this answer
    
isn't it $\left(\dfrac{3}{2}\right)^{n}$? –  Kamster Jul 13 at 21:11
    
Perhaps I've made a factoring error. $\frac{n+2^{n}}{n+3^{n}}=\frac{(\frac{n}{2^{n}}+1)2^{n}}{(\frac{n}{3^{n}}+1)3^{n‌​}}=\frac{1+\frac{n}{2^{n}}}{1+\frac{n}{3^{n}}}(\frac{2}{3})^{n}$? –  user71352 Jul 13 at 21:13
    
Oh no you are completely right, miss read it. I thought you took a $\frac{1}{2^{n}}$ and $\frac{1}{3^{n}}$ for some reason –  Kamster Jul 13 at 21:14
1  
Yea if there is any most common mistake I make in proofs, its algebra mistakes –  Kamster Jul 13 at 21:16
1  
I have the same problem. –  user71352 Jul 13 at 21:18

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.