Interval convergence of series help.

Question

What is the answer if the interval of the convergence of the series .

Answer 1

Use the Ratio Test:

lim_n→∞ l (a_n+1 / a_n l = lim_n→∞ [(n3ⁿ) / ((n+1)3ⁿ⁺¹] lx-2l = (1/3)lim_n→∞ [ (n/(n+1) lx-2l] = (1/3) lx-2l

By the Ratio Test, we get convergence when (1/3)lx-2l < 1.

So, -1 < (1/3)lx-2l < 1

-3 < x-2 < 3

-1 < x < 5

Check endpoints:

When x = 5, the series becomes ∑_{(n=1 to infinity)} (-1)ⁿ(1/n) which converges by the Alternating Series Test.

When x = -1, the series is ∑_{(n=1 to infinity)}(-1)²ⁿ(1/n) = ∑_{(n=1 to infinity)} (1/n) which is the Harmonic Series (divergent)

Interval of convergence is (-1, 5]

Radius of convergence = half the length of the interval of convergence = 3

Answer 2

Try to use the Ratio Test:

and take n->infinity. You can determine for which x it converges.

[P.S.: Typing on my phone. I will add more if get back to it from home]