You need to consider the number `A>1/5` such that:

`|a_n|<1/5<A => |a_n - 1/5|<A`

Substituting `((-1)^n)/(sqrt(n))` for `a_n` yields:

`|((-1)^n)/(sqrt(n)) - 1/5| < A`

Bringing the fractions to a common denominator yields:

`|(5*(-1)^n - sqrt n)/(5sqrtn)| < A`

Using the property of absolute value yields:

-`A < (5*(-1)^n - sqrt...

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You need to consider the number `A>1/5` such that:

`|a_n|<1/5<A => |a_n - 1/5|<A`

Substituting `((-1)^n)/(sqrt(n))` for `a_n` yields:

`|((-1)^n)/(sqrt(n)) - 1/5| < A`

Bringing the fractions to a common denominator yields:

`|(5*(-1)^n - sqrt n)/(5sqrtn)| < A`

Using the property of absolute value yields:

-`A < (5*(-1)^n - sqrt n)/(5sqrtn) < A`

Considering the inequality `(5*(-1)^n - sqrt n)/(5sqrtn) < A` yields:

`5*(-1)^n - sqrt n < 5Asqrt n`

`5*(-1)^n < 5Asqrt n + sqrt n`

Factoring out `sqrt n` yields:

`5*(-1)^n < sqrt n*(5A + 1) => (5*(-1)^n)/(5A + 1) < sqrt n`

Raising to square both sides yields:

`25/((5A+1)^2) < n`

**Hence, evaluating A, uner the given conditions, yields that there exists `n > 25/((5A+1)^2)` such that `|a_n|<1/5` .**