Question

Use the ratio test to determine whether the series is convergent or divergent.

Answer 1

Answer:

Option A is correct. since series is convergent.

Step-by-step explanation

nth term of the  given series is given by

$a_{n} =\frac{2n-1}{(2n-1)!}$

$a_{n+1} =\frac{2n+1}{(2n+1)!}$$\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)!}$

on simplifying it ,we get

$\lim_{n \to \infty}\frac{1}{(2n-1)(2n)}$

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 ]

Answer 2

Answer:

a) Convergent by ratio test

Step-by-step explanation:

Given is the series

$1+\frac{3}{1.2.3} +\frac{5}{1.2.3.4.5} +...$

General term =

$a_{n} =\frac{2n-1}{(2n-1)!}$

To use ratio test

Let us write n+1 th term

=$a_{n+1} =\frac{2n+1}{(2n+1)!}$

Find ratio of n+1th term to nth term

$\frac{a_{n+1} }{a_{n} } =\frac{2n+1}{2n-1} (\frac{1}{2n(2n+1} )$

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.