The series given is: 1 + 1 ! x </mrow> 2 </mfrac> +

Oakey1w

Oakey1w

Answered question

2022-06-25

The series given is:
1 + 1 ! x 2 + 2 ! x 2 3 2 + 3 ! x 3 4 3 + . . .

Answer & Explanation

Zayden Wiley

Zayden Wiley

Beginner2022-06-26Added 21 answers

Let u n = e n 1 n ! n n , n N . This is the same sequence as you except I did not simplified the n in the fraction.
u n + 1 u n = e n ( n + 1 ) ! ( n + 1 ) n + 1 × n n e n 1 n ! = e ( 1 + 1 n ) n = e exp ( n ln ( 1 + 1 n ) ) = e exp ( n ( 1 n 1 2 n 2 + o ( 1 n 2 ) ) ) = e exp ( 1 + 1 2 n + o ( 1 n ) ) = 1 + 1 2 n + o ( 1 n ) .
Therefore if n is big enough, u n + 1 u n 1 and ( u n ) is increasing. Since ( u n ) is positive, it means ( u n ) does not converges towards 0. Therefore the series ( u n ) diverges.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?