Question

Use the comparison test to determine whether the series is convergent or divergent

Answer 1

Answer:

b) diverges

Step-by-step explanation:

Given is a series with the terms

$\frac{25}{3} +\frac{125}{9} +\frac{625}{27} +...$

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

=$\frac{a(r^n-1}{r-1} =\frac{25}{3} (\frac{(\frac{5}{3} )^n-1}{\frac{5}{3}-1 } )$

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

Hence answer is

B) divergent

Answer 2

Answer:

Series is divergent .

Step-by-step explanation:

Here nth term of the given series  is

$\frac{5^{n+1} }{3^n}$

Let the given series be $a_{n} =$$\frac{5^{n+1} }{3^n}$

Let $b_{n} =$ = $\frac{5^{n} }{3^n}$

$\lim_{n \to \infty} \frac{a_n}{b_n} \\ \lim_{n \to \infty} \frac{\frac{5^(n+1)}{3^n} }{\frac{5^n}{3^n} }$

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  = $\frac{5^{n} }{3^n}$ is a geometric series with

                 common ratio $\frac{5}{3}$  >1

therefore for bn series is divergent.

therefore an also divergent ( using comparison test)