Answer:
Option b is correct.
Divergent
Comparison Test:
Let [tex]0\leq a_n\leq b_n[/tex] for all n.
If [tex]\sum_{n=1}^{\infty} b_n[/tex] converges, then [tex]\sum_{n=1}^{\infty} a_n[/tex] converges.
If [tex]\sum_{n=1}^{\infty} a_n[/tex] diverges, then [tex]\sum_{n=1}^{\infty} b_n[/tex] is also diverges.
Given the series [tex]\frac{25}{3} + \frac{125}{9} +\frac{625}{27}+....[/tex]
⇒[tex]a_n = \frac{5^{n+1}}{3^n}[/tex] for all natural number n.
or
[tex]a_n = 5(\frac{5^n}{3^n})[/tex]
Note that: [tex]4^n > 3^n[/tex] for all natural number n.
then;
[tex]\frac{1}{4^n}<\frac{1}{3^n}[/tex]
or
[tex]\frac{5^n}{4^n}<\frac{5^n}{3^n}[/tex]
[tex]5(\frac{5^n}{4^n})<5(\frac{5^n}{3^n})[/tex]
Geometric series:
[tex]\sum_{n=1}^{\infty} ar^n[/tex]
if [tex]|r| < 1[/tex], then the series is convergent.
if [tex]r\geq 1[/tex] then the series is divergent.
then;
by geometric series, [tex]\sum_{n=1}^{\infty} 5(\frac{5}{4})^n[/tex] it diverges as r > 1.
By comparison test:
[tex]\sum_{n=1}^{\infty} 5(\frac{5}{3})^n[/tex] diverges.
Therefore, the given series [tex]\frac{25}{3} + \frac{125}{9} +\frac{625}{27}+....[/tex] is divergent.
