Respuesta :
Answer:
Required solution gives series (a) divergent, (b) convergent, (c) divergent.
Step-by-step explanation:
(a) Given,
[tex]\sum_{n\to 0}^{\infty}\frac{2^n}{9^{2n}+1}[/tex]
To applying limit comparison test, let [tex]a_n=\frac{2^n}{9^{2n}+1}[/tex] and [tex]b_n=\frac{9^{2n}}{2^n}[/tex]. Then,
[tex]\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty}(1+\frac{1}{9^{2n}})=1>0[/tex]
Because of the existance of limit and the series [tex]\frac{9^{2n}}{2^n}[/tex] is divergent since [tex]\frac{9^{2n}}{2^n}=(\frac{9^2}{2})^n[/tex] where [tex]\frac{81}{2}>1[/tex], given series is divergent.
(b) Given,
[tex]\sum_{n\to 1}^{\infty}(\frac{7^n}{7^n+4})[/tex]
Again to apply limit comparison test let [tex]a_n=\frac{7^n}{7^n+4}[/tex] and [tex]b_n=\frac{1}{7^n}[/tex] we get,
[tex]\lim_{n\to \infty}\frac{a_n}{b_n}=\frac{1}{7^n+4}=0[/tex]
Since [tex]\lim_{n\to \infty} \frac{1}{7^n}=0[/tex] is convergent, by comparison test, given series is convergent.
(c) Given,
[tex]\sum_{n\to 1}^{\infty}\frac{5^n+2^n}{6^n}= \sum_{n\to 1}^{\infty}(\frac{5}{6})^n+\sum_{n\to 1}^{\infty}(\frac{1}{3})^n[/tex] . Now applying Cauchy Root test on last two series, we will get,
- \lim_{n\to \infty}|(\frac{5}{6})^n|^{\frac{1}{n}}=\frac{5}{6}=L_1
- \lim_{n\to \infty}|(\frac{1}{3})^n|^{\frac{1}{n}}=\frac{1}{3}=L_2
Therefore,
[tex]\lim_{n\to \infty}\frac{5^n+2^n}{6^n}=L_1+L_2=1.16>1[/tex]
Hence by Cauchy root test given series is divergent.