Calculus 2 Help! Series and Convergence?

Question

1. Find the sum. From n= 1 to infinity, of (4^n + 3^n)/12^n

2. Find the values of x for which the series  converges and find the sum of the series for those values of x.  n=0...infinity . 2^n(x+1)^n

3. Find the interval of convergence for the series. n=1...infinity. [(-2)^n(x+3)^n]/sqrt(n)

Please help. I've tried but failed. I am almost clueless when it comes to this convergence stuff. Answers would be great but explanations would be even better.

kb · Accepted Answer

1) This is the sum of two geometric series:

Σ(n = 1 to ∞) 4^n/12^n + Σ(n = 1 to ∞) 3^n/12^n

= Σ(n = 1 to ∞) (1/3)^n + Σ(n = 1 to ∞) (1/4)^n

= (1/3)/(1 - 1/3) + (1/4)/(1 - 1/4)

= 1/2 + 1/3

= 5/6.

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2) This is a geometric series with r = 2(x+1).

So, it converges <==> |2(x+1)| < 1 <==> |x + 1| < 1/2.

In which case, it converges to 1/(1 - 2(x+1)) = -1/(2x+1).

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3) Start with the Ratio Test.

r = lim(n→∞) |[(-2)^(n+1) (x+3)^(n+1) / √(n+1)] / [(-2)^n (x+3)^n / √n]|

..= 2|x+3| * lim(n→∞) √n/√(n+1)

..= 2|x+3| * lim(n→∞) √(n/(n+1))

..= 2|x+3| * √1

..= 2|x+3|.

So, the series converges when r = 2|x+3| < 1 <==> |x+3| < 1/2

and diverges when |x+3| > 1/2.

Now, we check the endpoints |x+3| = 1/2, or x = -5/2, -7/2:

x = -7/2 ==> Σ(n = 1 to ∞) 1/√n, divergent p-series (p = 1/2 < 1).

x = -5/2 ==> Σ(n = 1 to ∞) (-1)^n/√n, convergent by the Alternating Series Test, since {1/√n} decreases and converges to 0.

So, the interval of convergence is (-7/2, -5/2].

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I hope this helps!

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Calculus 2 Help! Series and Convergence?

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