I have the following series and want to prove it is converge using the squeezing theorem and root test

$$\sum_{i=1}^\infty \frac{(-1)^n + 5}{3^n}$$

  1. Just to bound it between two series and then use the root test on them?(If yes, then why should I do it when I can use simple comparison to one series)

  2. If I prove using the squeeze theorem for sequences that for $b_n < a_n < c_n$ then $\lim_{n\to\infty}a_n = 0$ can I say that my series $\sum{a_n}$ is convergent?

(of course I can prove it by series comparison to $ < \sum_{i=0}^\infty \frac{6}{3^n}$)



we now apply the squeeze theorem to get the limit of the $\;n\,-$ th root is less than one:

$$\frac13\xleftarrow[\infty\leftarrow n]{}\frac{\sqrt[n]4}3\le\frac{\sqrt[n]{(-1)^n+5}}3\le\frac{\sqrt[n]6}3\xrightarrow[n\to\infty]{}\frac13$$

  • $\begingroup$ Thanks , can u please answer to question 2 - what can i conclude about a series if I know that the sequence $an$ converge to zero? $\endgroup$ – Barak Mi May 14 '16 at 9:50
  • $\begingroup$ @BarakMi Nothing. You can't conclude anything at all. The condition $\;a_n\to0\;$ is a necessary one for convergence of the series, but it is far from being a sufficient condition. Observe that we did not do that here. We just evaluated the limit of $\;\sqrt[n]{|a_n|}\;$ by means of the squeeze theorem in order to use the $\;n\,-$ th root test. $\endgroup$ – DonAntonio May 14 '16 at 9:54

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