Question

Determine whether the series is absolutely convergent, conditionally convergent,or divergent.

\(\sum\limits_{n = 1}^\infty {\frac{{{\mathop{\rm Sin}\nolimits} 4n}}{{{4^n}}}} \)

Short answer

The Series is Absolutely Convergent.

Step-by-step solution

(Using Comparison Test)

\(1){u_n} \le {v_n},n \in {Z^ + },\sum {{v_n}} \)Converges then \(\sum {{u_n}} \)also Converges and Vice Versa for divergence.

Then Series is Absolutely Convergent if\(L < 1\),

Since,\({\mathop{\rm Sin}\nolimits} 4n \le 4n({\mathop{\rm Sin}\nolimits} \theta \le \theta ,\forall \theta )\)

=\(\frac{{{\mathop{\rm Sin}\nolimits} 4n}}{{{4^n}}} \le \frac{{4n}}{{{4^n}}} = {v_n}\)(Divided by \(4n\))

(Check For Convergence,By Ratio Test)

\({\lim _{n \to \infty }}|\frac{{{a_{n + 1}}}}{{{a_n}}}|\)=L

Then Series is Absolutely Convergent if\(L < 1\),

And Divergent When\(L > 1\)or\(\infty \)\(\)

\({\lim _{n \to \infty }}|\frac{{{a_{n + 1}}}}{{{a_n}}}|\)\( = {\lim _{n \to \infty }}|\frac{{4(n + 1)}}{{{4^{n + 1}}}}*\frac{{{4^n}}}{{4n}}|\)

=\(\frac{1}{4}.{\lim _{n \to \infty }}|\frac{{n(1 + \frac{1}{n})}}{n}|\)

=\(\frac{1}{4}\)

=\(\frac{1}{4} < 1\)

Hence,By Ratio Test\(\sum {{v_n}} \)is Convergent..

So,By Comparison Test\(\sum {{u_n}} \)also converges\(\)..

Therefore,\(|\sum {{u_n}|} \)Converges.

Hence,The Given Series is Absolutely Convergent..