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Zanzabum
3 years ago
9

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

Mathematics
2 answers:
Lunna [17]3 years ago
4 0

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


frez [133]3 years ago
4 0

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)

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