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&l…

Question

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.

Answer 1

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.