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Use the WeierstrassM-test to show that each of the following series converges uniformly on the given domain:


\sum _(k\geq0) (z^k)/(z^k+1) on the domain
\bar{D} [0,r], where 0≤r<1.

1 Answer

3 votes

Answer:

Explanation:

Given a series
\sum_(k=1)^\infty f(z), the Weierstrass M-test tell us that if we find a sequence of positive numbers
M_n such that
|f(z)|\leq M_n in a certain domain D, and the series
\sum_(n=1)^\infty M_k converges, then the series
\sum_(k=1)^\infty f_k(z) converges uniformly in the domain D.

So, our objective is to find the so called sequence
M_k. The main idea is to bound the sequence of functions
(z^k)/(z^k + 1).

Now, notice that the values of z are always positive, so
z^k is always positive, so
z^k+1\geq 1 for all values of z in
\overline{D}. Then,


\Big| (z^k)/(z^k + 1)\Big| \leq z^k,

because if we make the values of the denominator smaller, the whole fraction becomes larger.

Moreover, as z is in the interval [0,r], we have that
z\leq r and as consequence
|z^k|\leq r^k. With this in addition to the previous bound we obtain


\Big| (z^k)/(z^k + 1)\Big| \leq |z^k|\leq r^k.

With this, our sequence is
M_k = r^k and the corresponding series is
\sum_(k=1)^\infty r^k, which is a geometric series with ratio less than 1, hence it is convergent.

Then, as consequence of Weierstrass M-test we have the uniform convergence of the series in the given domain.

User Iivannov
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