###### Question:

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

## Answers

Step-by-step explanation:

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

So, our objective is to find the so called sequence [tex]M_k[/tex]. The main idea is to bound the sequence of functions [tex]\frac{z^k}{z^k + 1}[/tex].

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

[tex]\Big| \frac{z^k}{z^k + 1}\Big| \leq z^k,[/tex]

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 [tex]z\leq r[/tex] and as consequence [tex]|z^k|\leq r^k[/tex]. With this in addition to the previous bound we obtain

[tex]\Big| \frac{z^k}{z^k + 1}\Big| \leq |z^k|\leq r^k.[/tex]

With this, our sequence is [tex]M_k = r^k[/tex] and the corresponding series is [tex]\sum_{k=1}^\infty r^k[/tex], 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.

[tex]a_k=\left|\dfrac{k^2-3}{k^2+4}\right|;\text{ is nondecreasing for $k2$}[/tex]

[tex]\lim\limits_{k \to \infty} a_k =1[/tex]

The series does not converge

Step-by-step explanation:

Given ...

[tex]S=\displaystyle\sum\limits_{k=2}^{\infty}{x_k}\\\\x_k=(-1)^k\cdot\dfrac{k^2-3}{k^2+4}\\\\a_k=|x_k|[/tex]

Find

whether S converges.

Solution

The (-1)^k factor has a magnitude of 1, so the magnitude of term k can be written as ...

[tex]\boxed{a_k=1-\dfrac{7}{k^2+4}}[/tex]

This is non-decreasing for k>1 (all k-values of interest)

As k gets large, the fraction tends toward zero, so we have ...

[tex]\boxed{\lim\limits_{k\to\infty}{a_k}=1}[/tex]

Terms of the sum alternate sign, approaching a difference of 1. The series does not converge.

The given series neither converges nor diverges.

Step-by-step explanation: We are given to determine whether the following series converges or diverges :

[tex]S=\sum_{n=0}^{\infty}(-1)^n.[/tex]

If the series converges, we are to find its sum.

The given series can be written as :

[tex]1,~-1,~1,~-1,~1,~1,~~.~~.~~.[/tex]

We note that the given series is a geometric one with first term 1 and common ratio given by

[tex]r=\dfrac{-1}{1}=\dfrac{1}{-1}=~~.~~.~~.~~=-1.[/tex]

We know that a geometric series with common ratio r converges if |r| <1 and diverges if |r| > 1.

Since |r| = 1 for the given series, so the series will neither converge nor diverge.

Thus, the given series neither converges nor diverges.

1st answer is DIV 2nd answer is MNIF and 3rd answer is DIV

Step-by-step explanation:

1. ∑n=1[infinity](sin(10n)−sin(10n+1))

By applying formula sin C - sin D =2sin(C-D)/2cos(C+D)/2∑n=1[infinity](2sin(-1/2)cos(10n+1/2)By expanding the summation we get -2sin 1/2[ cos(10+1/2)+cos(20+1/2)+cos(30+1/2)+cos(40+1/2)+cos(50+1/2)+cos(60+1/2)+cos(70+1/2)+cos(80+1/2)-sin(1/2)-sin(10+1/2)-sin(20+1/2)-sin(30+1/2)-sin(40+1/2)-sin(50+1/2)-sin(60+1/2)-sin(70+1/2)-sin(80+1/2)-cos(1/2)-cos(10+1/2)-cos(20+1/2)sin(∞ )]All the term cancels out and sin(∞) and other last term is left out which value we don't know so the answer is DIV2. ∑n=1[infinity](e∧11n−e∧11(n+1))

By expanding the summation we get= e∧11-e∧22+e∧22-e∧33+e∧33-e∧44-e∞=e∧11-e∞e∞>>>>>>>e∧11=-∞So the answer is MNIF3. ∑n=1[infinity](sin(10n)−sin(10(n+1)))

By expanding the summation we get=sin(10)-sin(20)+sin(20)-sin(30)+sin(30)-sin(40)-sin(∞)=sin(10)-sin(∞)sin(∞) value we don't know so we can't decide the answer so the answer DIVSince the limit resulting from the ratios test is = 1 / 729 which is < 1 the series converges absolutely

Step-by-step explanation:

Since the limit resulting from the ratios test is = 1 / 729 which is < 1 the series converges absolutely

ATTACHED BELOW IS THE DETAILED SOLUTION of the above answer

[tex]Use the Ratio Test to determine if the following series converges absolutely or diverges. sigma^inf[/tex]

An alternating series [tex]\sum\limits_n(-1)^na_n[/tex] converges if [tex]|(-1)^na_n|=|a_n|[/tex] is monotonic and [tex]a_n\to0[/tex] as [tex]n\to\infty[/tex]. Here [tex]a_n=\dfrac1{\ln(n+1)}[/tex].

Let [tex]f(x)=\ln(x+1)[/tex]. Then [tex]f'(x)=\dfrac1{x+1}[/tex], which is positive for all [tex]x-1[/tex], so [tex]\ln(x+1)[/tex] is monotonically increasing for [tex]x-1[/tex]. This would mean [tex]\dfrac1{\ln(x+1)}[/tex] must be a monotonically decreasing sequence over the same interval, and so must [tex]a_n[/tex].

Because [tex]a_n[/tex] is monotonically increasing, but will still always be positive, it follows that [tex]a_n\to0[/tex] as [tex]n\to\infty[/tex].

So, [tex]\sum\limits_n(-1)^na_n[/tex] converges.

It is converge because it is an infinite sequence

The common ratio is:

-21/7=63/-21=-3

So r=-3

A series only converges if r^2<1, since r^2 in this case is 9, the series diverges.