Show the sequence is divergent

Question

Hi,

I need to show that the following sequence is divergent,

n³cos(πn)/n²+1, n=1,2,...

I know i need to use the recipital rule so n³cos(πn)/n²+1 becomes n²+1/n³cos(πn) and then i need to show this is converget to 0.

I'm not sure whta i do with the trig function, as it converges between 1 and -1.

Thnaks Kara

Answer 1

Continuing from where you left off:

(n²+1)/n³cos(πn)

We need to show that this converges to 0. It is a theorem that a sequence converges to 0 if and only if its absolute value converges to 0. As you noted, cos(πn) alternates between 1 and -1 for n an integer. Then its absolute value is always just 1.

So (n²+1)/n³cos(πn) converges to 0 if and only if (n²+1)/n³ converges to 0.

But (n²+1)/n³ = n²/n³ + 1/n³ = 1/n + 1/n³. The limit of a sum is the sum of the limits and these both tend to 0, so we are done.

Answer 2

Hi Kara,

You are right on track in knowing that you can show that the reciprocal sequence converges to 0.

The only thing the cos(nπ) part of the expression is doing is causing the terms to alternate signs, + to -. (cosπ = -1 , cos2π = 1, cos3π = -1, etc.) (Btw, most alternating sequences just use (-1)ⁿ in the expression, so the cos part is just a little trickier than necessary.)

So, analyzing the reciprocal sequence ignoring the sign changes (in other words, looking at the absolute value of the terms in the reciprocal sequence) we have t_n = (n² + 1) / n³ , which converges to 0 because the exponent in the denominator is bigger than the exponent in the numerator. A formal way to show this is to separate the expression into two fractions by dividing each term of the numerator by n³, giving us 1/n + 1/n³, both of which converge to 0.

Finally, since the reciprocals converge to 0, the original function diverges.

Answer 3

Let find lim n→∞ n³cos(nπ)/(nπ)/(n²+ 1)= lim n→∞ (- 1)ⁿn³/(n² + 1) DNE,

because this sequence is: - 1/2, 8/5, - 27/10, ...