How to determine series

Question

Let f(x)= ∑_n=1^∞ (-1)^n (x/4)^n. Find the series and interval of convergence for the integral from 0 to x of f(t)dt

Answer 1

Hi Bob,

I hope this is helpful.

First, let's simplify the function a little bit

f(x) = ∑_n=1^∞ [(-1)ⁿ * xⁿ ] / 4ⁿ .

Then we can use the ratio test;

lim_n->∞ |a_n+1 / a_n| (1)

If (1) < 1, the series converges. If (1) > 1, the series converges. If (1) = 0, it is inconclusive.

Now, you want to find the interval of convergence. To find that we need (1) < 1.

lim_n->∞ | [{(-1)ⁿ⁺¹ * xⁿ⁺¹} / 4ⁿ⁺¹] / [{(-1)ⁿ * xⁿ} / 4ⁿ] | = lim_n->∞ | [{(-1)ⁿ⁺¹ * xⁿ⁺¹} / 4ⁿ⁺¹] * [4ⁿ / {(-1)ⁿ * xⁿ}] |

= lim_n->∞ | [(-1) * x] / 4 | = | -x / 4 | < 1 (2)

Now, solve (2) for x. Then you get -1 < -x / 4 < 1 ⇒ -4 < -x < 4 ⇒ -4 < x < 4.

Thus, we know that the series converges when x = (-4,4).

Finally, let's check the points, where (1) = 1.

| -x / 4 | = 1 then, -x / 4 = 1 or -x / 4 = -1.

Thus, x = 4 or -4.

Let's check x = 4 first. Then f(4) = ∑_n=1^∞ [(-1)ⁿ * 4ⁿ] / 4ⁿ = ∑_n=1^∞ (-1)ⁿ.

Now, apply the divergence test to f(4) and you will find out that the series diverges at x = 4.

For the same reason, the series also diverges at x = -4.

Therefore, the interval of convergence for this series is -4 < x < 4.

**Note; Divergence Test**--------------------

If lim_n->∞ a_n ≠ 0, ∑_n=1^∞ a_n ⇒ Diverges |

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Now I am realizing that I may have not answered your question, but here is what I have for the interval of convergence of the given series.

Best

++ ADDITION++

The integral from 0 to x f(t)dt

∫₀^x ∑_n=1^∞ [(-1)ⁿ * tⁿ] / 4ⁿ dt = ∑_n=1^∞ [(-1)ⁿ * tⁿ⁺¹] / [(n+1) * 4ⁿ] |₀^x = ∑_n=1^∞ [(-1)ⁿ * xⁿ⁺¹] / [(n+1) * 4ⁿ].

This is now your new series and you can use the same test above to find the interval of convergence.

Best,

Sunghyun