Using the squeeze theorem, show that the series converges and it's limit.
(-1)^n(3/2^n)
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.
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!
Comments
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)).
------------------
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!