1) Show the power series: å¥ n = 0 z n
for which the series converges provided z < 1.
(I.e. if ÷ z n
+1 /
z n ÷ < k, for all k sufficiently large and k < 1
then:
å z n
converges absolutely.
For example, expanding the series terms:
å¥ n = 0 z n
= 1 + z + z2 + z3 +…..
Let z = ½ :
å¥ n = 0 z n
= 1 + ½
+ (½) 2 + (½) 3
= 1 + ½ + 1/4 + 1/8 +
……
So since the sum is limiting as terms are added, not expanding,
we see the series converges absolutely.
BUT this only obtains for all points INSIDE a circle of radius R =1, so
the radius of convergence is R =1.
2)Consider the power series:
å¥ n = 0 (z) n / n!
Show that the radius of convergence R = ¥
Apply the ratio test:
lim n ® ¥
[(z) n
+1 / n + 1 / (z) n / n ] =
lim n ® ¥
a n / a n +1 = n + 1 ® ¥
as n tends to infinity, hence R = ¥
3) The Maclaurin series for the normal exponential function
f(z) = exp(z) is:
å¥ n = 0 (z) n / n! = 1 + z + z2
/ 2! + z3 / 3! +…..
If we now replace z by (-z) we get:
å¥ n = 0 (-z) n / n! = 1 - z + z2 / 2! - z3 / 3! +…..
Which is the Maclaurin series for exp(-z)
Then for radius of
convergence: R = 1/L = 1/0
= ¥
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