1) Show the power series: å

^{¥}_{n = 0}z_{n }_{ n ® ¥}[(z)

^{n +1}^{ }

_{ }

**/**(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 + z^{2}+ z^{3}+…..
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 + z^{2}/ 2! + z^{3}/ 3! +…..
If we now replace z by (-z) we get:

å

^{¥}_{n = 0}(-z)^{n}_{ }**/**n! = 1 - z + z^{2}/ 2! - z^{3}/ 3! +…..
Which is the Maclaurin series for exp(-z)

**Then for radius of convergence: R = 1/L = 1/0 =**¥

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