Homework Help

What is limit in sequence a(subscript n)=n^3ln(n+2/n+1)ln(n+3/n+2)ln(n+4/n+3)?

user profile pic

greenbel | Honors

Posted August 20, 2013 at 2:23 PM via web

dislike 1 like

What is limit in sequence a(subscript n)=n^3ln(n+2/n+1)ln(n+3/n+2)ln(n+4/n+3)?

1 Answer | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 20, 2013 at 2:44 PM (Answer #1)

dislike 0 like

You need to re-write the general term of series `a_n` , such that:

`a_n = n*ln((n+2)/(n+1))*n*ln((n+3)/(n+2))*n*ln((n+4)/(n+3))`

Using the power property of logarithms yields:

`a_n = (ln((n+2)/(n+1))^n)*(ln((n+3)/(n+2))^n)*(ln((n+4)/(n+3))^n)`

Taking the limit, yields:

`lim_(n->oo) a_n = lim_(n->oo) (ln((n+2)/(n+1))^n)*(ln((n+3)/(n+2))^n)*(ln((n+4)/(n+3))^n)`

`lim_(n->oo) a_n = lim_(n->oo) (ln((n+2)/(n+1))^n)*lim_(n->oo) (ln((n+3)/(n+2))^n)*lim_(n->oo) (ln((n+4)/(n+3))^n)`

Taking each limit separately, yields:

`lim_(n->oo) (ln((n+2)/(n+1))^n) = ln lim_(n->oo) ((n+2)/(n+1))^n`

Using the sepcial limit `lim_(n->oo) (1 + 1/n)^n = e` , yields:

`lim_(n->oo) ((n+2)/(n+1))^n = lim_(n->oo) (1 + (n+2)/(n+1) - 1)^n`

`lim_(n->oo) (1 + (n + 2 - n - 1)/(n+1))^n = lim_(n->oo) (1 + 1/(n+1))^n`

`lim_(n->oo) (1 + 1/(n+1))^n = lim_(n->oo) ((1 + 1/(n+1))^(n+1))^(n/(n+1))`

Since `lim_(n->oo) ((1 + 1/(n+1))^(n+1)) = e` , yields:

`lim_(n->oo) (1 + 1/(n+1))^n = e^(lim_(n->oo)(n/(n+1))) = e^1`

`lim_(n->oo) (1 + 1/(n+1))^n = e`

Replacing e for `lim_(n->oo) ((n+2)/(n+1))^n` yields:

`lim_(n->oo) (ln((n+2)/(n+1))^n) = ln e = 1`

You need to evaluate the next limit, such that:

`lim_(n->oo) (ln((n+3)/(n+2))^n) = ln lim_(n->oo) ((n+3)/(n+2))^n`

Reasoning by analogy, yields:

`lim_(n->oo) (ln((n+3)/(n+2))^n) = ln e = 1`

Evaluating the last limit, yields:

`lim_(n->oo) (ln((n+4)/(n+3))^n) = ln e = 1`

`lim_(n->oo) a_n = 1*1*1 = 1`

Hence, evaluating the limit of the given series a_n, using the special limit `lim_(n->oo) (1 + 1/n)^n = e` , yields `lim_(n->oo) a_n = 1` .

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes