Respuesta :

Answer:

a) Convergent by ratio test


Step-by-step explanation:

Given is the series

[tex]1+\frac{3}{1.2.3} +\frac{5}{1.2.3.4.5} +...[/tex]

General term =

[tex]a_{n} =\frac{2n-1}{(2n-1)!}[/tex]

To use ratio test

Let us write n+1 th term

=[tex]a_{n+1} =\frac{2n+1}{(2n+1)!}[/tex]

Find ratio of n+1th term to nth term

[tex]\frac{a_{n+1} }{a_{n} } =\frac{2n+1}{2n-1} (\frac{1}{2n(2n+1} )[/tex]

We find that here numerator has degree as 1 and denominator as 3

SInce numerator has more powers, it approaches infinity faster than numerator thus making ratio tend to 0 as n becomes large

Since ratio tends to 0, we have this series is convergent.

Answer:

Option A is correct. since series is convergent.

Step-by-step explanation

nth term of the  given series is given by

[tex]a_{n} =\frac{2n-1}{(2n-1)!}[/tex]

[tex]a_{n+1} =\frac{2n+1}{(2n+1)!}[/tex][tex]\lim_{n \to \infty} \frac{a_(n+1)}{a_{n} } =\frac{\frac{2n+1}{2n+1!} }{\frac{2n-1}{2n-1!}  } =\frac{(2n+1)X(2n-1)!}{(2n-1)X(2n+1)!}[/tex]

on simplifying it ,we get

[tex]\lim_{n \to \infty}\frac{1}{(2n-1)(2n)}[/tex]

which gives zero at n = infinity

since value of the limit of the ratio is less than 1

given series is convergent  [ by ratio test ]


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