Calculus series problem. Please help!?

bilal-r

Using the squeeze theorem, show that the series converges and it's limit.

(-1)^n(3/2^n)

kb

Assuming that the index of summation starts at n = 0,

consider S = Σ(n=0 to k) 3(-1/2)^n = 3 - 3/2 + 3/4 - ... + 3(-1/2)^k.

Multiply both sides by (-1/2):

-S/2 = Σ(n=0 to k) 3(-1/2)^(n+1) = -3/2 + 3/4 - ... + 3(-1/2)^k + 3(-1/2)^(k+1).

Subtract the two series:

S - (-S/2) = 3 - 3(-1/2)^(k+1)

==> 3S/2 = 3(1 - (-1/2)^(k+1))

==> S = 2(1 - (-1/2)^(k+1)).

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Now, we show by the Squeeze Law that lim(k→∞) (-1/2)^(k+1) = 0.

Note that -(1/2)^(k+1) ≤ (-1/2)^(k+1) ≤ (1/2)^(k+1) for all k.

Since lim(k→∞) ±(1/2)^(k+1) = 0, the result follows.

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Finally, Σ(n=0 to ∞) 3(-1/2)^n

= lim(k→∞) 2(1 - (-1/2)^(k+1))

= 2(1 - 0)

= 2.

I hope this helps!