Question

test the convergence of the series:- 1+ 3/2x+5/9x²+7/28x³+9/65x⁴+.......​

Answer 1

The given series will converge if |x| < 1, and it will diverge if |x| > 1.

To test,

Convergence of the series.

Given,

1+ 3/2x+5/9x²+7/28x³+9/65x⁴+.......​

Solution,

We can test the convergence of the series using the ratio test:

lim (n->∞) |(a_{n+1}/a_n)|

where a_n is the nth term of the series.

Here, the nth term of the series is given by:

a_n = (2n-1)/(n²+1) x^n

Therefore, the (n+1)th term of the series is given by:

a_{n+1} = (2(n+1)-1)/((n+1)²+1) x^{n+1} = (2n+1)/((n+1)²+1) x^{n+1}

Taking the ratio of the (n+1)th and nth terms, we get:

a_{n+1}/a_n = [(2n+1)/((n+1)²+1) x^{n+1}] / [(2n-1)/(n²+1) x^n]

= [(2n+1)/(2n-1)] x [(n²+1)/((n+1)²+1)] x x

= [(2n+1)/(2n-1)] x [(n^2+1)/(n^2+2n+2)] x x

As n approaches infinity, the term x^n dominates the expression, and we can ignore the other factors. Therefore, we have:

lim (n->∞) |(a_{n+1}/a_n)| = lim (n->∞) |[(2n+1)/(2n-1)] x [(n^2+1)/(n^2+2n+2)] x|

= lim (n->∞) |(2n+1)/(2n-1)| x lim (n->∞) |(n^2+1)/(n^2+2n+2)| x |x|

= 1 x 1 x |x|

= |x|

The series will converge if |x| < 1, and it will diverge if |x| > 1.

Therefore, we can conclude that the given series will converge if |x| < 1, and it will diverge if |x| > 1. If |x| = 1, then the ratio test is inconclusive and we need to use another method to determine convergence or divergence.

For further reference visit,

https://brainly.in/question/20598098?referrer=searchResults

https://brainly.in/question/25699077?referrer=searchResults

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