Respuesta :

Answer:

b) diverges

Step-by-step explanation:

Given is a series with the terms

[tex]\frac{25}{3} +\frac{125}{9} +\frac{625}{27} +...[/tex]

We find that this series follows a pattern such that each term is multiplied by 5/3 to get the next term.

In other words, this is a geometric series with  I term=25/3 and common ratio=5/3

Sum of geometric series upto n terms

=[tex]\frac{a(r^n-1}{r-1} =\frac{25}{3} (\frac{(\frac{5}{3} )^n-1}{\frac{5}{3}-1 } )\\[/tex]

Since r>1, we find that this series diverges to infinity

Hence answer is

B) divergent

Answer:

Series is divergent .

Step-by-step explanation:

Here nth term of the given series  is

[tex]\frac{5^{n+1} }{3^n}[/tex]

Let the given series be [tex]a_{n} =[/tex][tex]\frac{5^{n+1} }{3^n}[/tex]

Let [tex]b_{n} =[/tex] = [tex]\frac{5^{n} }{3^n}[/tex]

[tex]\lim_{n \to \infty} \frac{a_n}{b_n} \\ \lim_{n \to \infty} \frac{\frac{5^(n+1)}{3^n} }{\frac{5^n}{3^n} }[/tex]

on simplifying it ,we get

limit of the ratio  = 5 (which is finite and nonzero)

therefore using comparison  test both an and bn converge or diverge together .

for bn  = [tex]\frac{5^{n} }{3^n}[/tex] is a geometric series with

                 common ratio [tex]\frac{5}{3}[/tex]  >1

therefore for bn series is divergent.

therefore an also divergent ( using comparison test)

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