# Convergence of: $\sum_{n=1}^{\infty} \frac{2+(-1)^n}{2^n+(-1)^n}$

I've tried solving this exercise but got stuck on a big expression that I could not untangle. I've obtainded the following thing: $$\lim_{n \to \infty} n\frac{2 \cdot 2^n +3 \cdot 2^n\cdot (-1)^n-4\cdot (-1)^n}{(2^n+(-1)^n)(2+(-1)^{n+1})}$$

I used Raabe-Duhamel theorem, after seeing that using the ratio test is too difficult.

The series is: Determine the nature of: $$\sum_{n=1}^{\infty} \frac{2+(-1)^n}{2^n+(-1)^n}$$

In my textbook is required to use ratio test or Raabe-Duhamel. Can you please help me ?

• Hint: $2^n+(-1)^n> 2^{n-1}$. Please try to be not too cryptic (or improve your English): determine the nature has not mathematically meaning IMHO. – Karl Jan 20 '15 at 14:10
• I'm not a native english speaker, I'm trying to write everything as well as I can, sorry – Ivan Gandacov Jan 20 '15 at 14:17

As noted by Karl $$\underset{n\geq1}{\sum}\frac{2+\left(-1\right)^{n}}{2^{n}+\left(-1\right)^{n}}<3\underset{n\geq1}{\sum}\frac{1}{2^{n-1}}=3\underset{n\geq0}{\sum}\frac{1}{2^{n}}=6.$$
• Thank you, but how did you get: $3\sum_{n}^{\infty} \frac{1}{2^{n-1}}$. The denominator is clear, but the numerator ? – Ivan Gandacov Jan 20 '15 at 15:25
• Because $$2+\left(-1\right)^{n}=\begin{cases} 3, & n\, even\\ 1, & n\, odd \end{cases}.$$ – Marco Cantarini Jan 20 '15 at 15:26