sequences converge/diverge

Question

Determine whether the following sequences converge or diverge. Justify your answers (show why)

1. a_n = (-1)⁽²ⁿ⁺¹⁾
2. a_n = ln[(e⁴)ⁿ]/(3n)
3. a_n = (eⁿ + e^-n)/(e²ⁿ-1)

Answer 1

1: a_n = (-1)⁽²ⁿ⁺¹⁾

This may look like an alternating series, but it isn't. Since there is a term of 2n in the exponent in the explicit formula, each value of this series is -1. This is therefore a constant sequence, which converges to that constant.

a₀ = (-1)¹ = -1

a₁ = (-1)³ = -1

a₂ = (-1)⁵ = -1

...

2: a_n = ln[(e⁴)ⁿ]/(3n)

lets simplify using the power rule for logarithms

ln[(e⁴)ⁿ]/(3n =

n*ln(e⁴)/(3n) =

4n / (3n) =

(4/3)

Another constant sequence! (convergent)

3: a_n = (eⁿ + e^-n)/(e²ⁿ-1)

Both the numerator and the denominator in this will approach infinity, but that does not mean that they will approach it equally as quickly, so we do not yet know if this sequences converges.

Lets divide the numerator and the denominator by eⁿ so the denominator will not approach infinity

(eⁿ + e^-n)/(e²ⁿ-1) =

(1 + e^-2n)/ (eⁿ - e^-n)

the limit of e^-n as n approaches infinity is 0. The same is true for e^-2n. Our limit can therefore be simplified:

lim [(1 + e^-2n)/ (eⁿ - e^-n)] =

n->∞

lim [1/ eⁿ ]

n->∞

1/eⁿ = e^-n approaches 0

Therefore a_n = (eⁿ + e^-n)/(e²ⁿ-1) converges as n approaches infinity. It does not converge at negative infinity.