# What is a, if given series`(sqrt((1-a^2)^2*n^2+2n))(1/n)` converge to 3 if n unlimited?

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### 1 Answer

You need to determine the unknown "a" if the limit of the series `sqrt((1-a^2)^2*n^2+2n)/n` is 3, if n goes to `oo` .

Solving the limit of the series yields:

`lim_(n-gtoo) (sqrt (n^2*((1-a^2)^2 + 2/n)))/n = lim_(n-gtoo) (n*sqrt ((1-a^2)^2 + 2/n))/n`

Reducing by n yields:

`lim_(n-gtoo) (sqrt ((1-a^2)^2 + 2/n))= `

`= lim_(n-gtoo) sqrt (1-a^2)^2) + lim_(n-gtoo) (2/n) = (sqrt ((1-a^2)^2) + 0`

This limit is `3 =gt (sqrt ((1-a^2)^2) = 3 =gt |1-a^2| = 3`

Using the property of absolute value yields:

`1 - a^2 = 3 =gt a^2 = 1-3 =gt a^2 = -2 =gt` there is no real values for "a" such that `a^2 = -2` .

`1 - a^2 = -3 =gt a^2 = 1+3 =gt a^2 = 4 =gt a = +-2`

**The values of the unknown "a" such that `lim_(n-gtoo) (sqrt (n^2*((1-a^2)^2 + 2/n)))/n = 3` are: `a = +-2` .**