What is the interval of convergence of the infinite series (below)?

Question

oo
∫ [k^2 / (k!)](x + 1)^k.
k=0


All the possible solutions given by the book. Which is correct?

(2, 4)

[2, 4)

(2, 4]

[-1/3, 1/3)

(-1/3, 1/3)

[-1/3, 1/3]

(-1, 1]

[-1, 1)

[-1, 1]

(-2, 2)

(-4, 4)

x = 0 only

x = 2 only

x = 4 only

(-oo, oo)

or is it none of these

Answer 1

(-∞, ∞)

Here's why:

By the ratio test, if [k→∞]lim |f(k+1)/f(k)| < 1, then [k=0, ∞]∑f(x) converges absolutely. Now, we apply that to this series:

[k→∞]lim |((k+1)²(x+1)^(k+1) / (k+1)!)/(k²(x+1)^k/k!)|

Simplifying a bit:

[k→∞]lim |(k+1)²/k² (x+1)/(k+1)|

Which we write as the product of two limits:

[k→∞]lim |(k+1)²/k²| * [k→∞]lim |(x+1)/(k+1)|

We see easily that the limit on the left is 1, and that the limit on the right is 0, so indeed we have:

[k→∞]lim |((k+1)²(x+1)^(k+1)/(k+1)!) / (k²(x+1)^k/k!)| = 1*0 = 0 < 1

So the series converges absolutely. Note that this is true _regardless of the value of x_, so in fact this series converges on the entire real line, which is (-∞, ∞).