∑(((x-9)^n)/(9+7n)) n=0 to infinity. Find interval of convergence?

Instead of the 'x', did you mean 'n'?

EDIT:

Oh, I see. This is a Power Series.

You might have not been using the Ratio Test properly.

Applying it, yields:

lim n-->oo | [(x-9)^(n+1) / (9+7(n+1))] / [(x-9)^n / (9+7n)] |,

lim n-->oo | [(x-9)^(n+1) / (9+7(n+1))] * [(9+7n) / (x-9)^n] | ((x-9)^n on top cancels),

lim n-->oo | [(x-9)(9+7n)] / (16+7n) |,

|x-9| lim n-->oo | (9/n + 7) / (16/n + 7) | = 1.

Therefore our limit is 1.

From Power Series, we know that:

|x-9| < 1, series converges;

|x-9| > 1, series diverges.

We get that our radius of convergence in this case is just R = 1.

Now, to find the interval of convergence, let's solve the inequality.

|x-9| < 1

-1 < x - 9 < 1

8 < x < 10.

Finally, let's plug in each of our endpoints in the series.

First x = 8. Doing so, gives:

oo

Σ (8-9)^n / (9+7n)

n=0,

oo

Σ -1^n / (9+7n)

n=0,

Notice that this is in the form of the Alternating Harmonic series.

Therefore, this first sum converges.

Now for our other endpoint.

oo

Σ (10-9)^n / (9+7n)

n=0,

oo

Σ 1^n / (9+7n)

n=0,

oo

Σ 1 / (9+7n)

n=0.

For this next sum, let's use the integral test.

Doing so, yields:

oo

∫ 1 / (9+7x) dx

0

Let, u = 9 + 7x, du = 7 dx --> du/7 = dx,

1/7 ∫ u^-1 du,

(1/7)ln| 9 + 7x | [ (0,c),

(1/7)ln| 9 | - (1/7)ln| 9 + 7c |.

Taking the limit, gives:

(1/7) lim c-->oo ln | 9 | - ln| 9 + 7c |,

(1/7) lim c-->oo ln | 9 / (9 + 7c) |,

(1/7) lim c-->oo ln | 9 / oo | = ln | 0 | = -oo.

Therefore, this sum diverges.

Finally, we conclude that our interval of convergence is:

8 ≤ x < 10.

Hope this helped.