# Determine if the alternating series converges absolutely, conditionally or diverges

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. 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.

• 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 '14 at 22:02
• 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 '14 at 22:06
• You are welcome. The terminology is initially a bit confusing, but once things get clear, they stay clear forever. – André Nicolas Jul 13 '14 at 22:47

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

• isn't it $\left(\dfrac{3}{2}\right)^{n}$? – Kamster Jul 13 '14 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 '14 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 '14 at 21:14
• Yea if there is any most common mistake I make in proofs, its algebra mistakes – Kamster Jul 13 '14 at 21:16
• I have the same problem. – user71352 Jul 13 '14 at 21:18