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 ?

  • 4
    $\begingroup$ 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. $\endgroup$ – Karl Jan 20 '15 at 14:10
  • $\begingroup$ I'm not a native english speaker, I'm trying to write everything as well as I can, sorry $\endgroup$ – Ivan Gandacov Jan 20 '15 at 14:17

As noted by Karl $$\underset{n\geq1}{\sum}\frac{2+\left(-1\right)^{n}}{2^{n}+\left(-1\right)^{n}}<3\underset{n\geq1}{\sum}\frac{1}{2^{n-1}}=3\underset{n\geq0}{\sum}\frac{1}{2^{n}}=6.$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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