## prove that a sequence whose range is relatively compact admits a convergent partial sequence

I need help with this demo please.
prove that a sequence whose range is relatively compact admits a convergent partial sequence.
the truth is that I have looked for information but I can not

## Used the definition of convergent proof \$z_n\$ convergent

Let $$x_{n}$$ converges to a and $$y_{n}$$ converges to b. Then prove that

$$z_{n} = frac{2bx_{n}+2ay_{n} – 2x_{n}y_{n} + ab}{3}$$

Proof that $$z_n$$ convergent to $$ab$$

## Is it possible for a power series to be conditionally convergent at two different points?

Like I stated in the title, I was just wondering if it’s possible for a power series to be conditionally convergent at two different points. Are there any examples of power series that fit this criteria?

Any help is appreciated!

## real analysis – Comparing divergent and convergent sums

Let $$(x_n)$$ be a monotonically decreasing sequence of positive real numbers that is also summable.

Let $$(y_n)$$ be a sequence of positive real numbers such that
$$sum_n x_n y_n$$ converges.

Let $$(z_n)$$ be a monotonically increasing sequence of positive real numbers such that $$sum_n x_n z_n =infty.$$

Assume that the sequences $$y_n$$ and $$z_n$$ are such that $$2^{-varepsilon y_n}$$ and $$2^{-varepsilon z_n}$$ are summable for every $$varepsilon>0.$$ Does it follow that there is some $$delta>0$$ such that

$$sum_n Big(2^{-varepsilon y_n}-2^{-varepsilon z_n}Big) ge 0 text{ for all } varepsilon in (0,delta)?$$

The motivation for this statement to be true is that $$z_n$$ should be larger most of the time than $$y_n$$ and we capture this most of the time by taking $$varepsilon$$ small.

## Prove SEIRS series is convergent

I would like to prove that the SEIRS series:

$$S^{n+1} = 1-I + frac{β}{α}ln(S^{n})$$

where:
$$S^{0}=1$$ and I,β,α are constants.

is convergent as long as a*s<β

## NIntegrate of a convergent integral working with large integration limits, but not with infinite integration limits

Code for reproducing is below:

``````Integrand(x_) := ((x + 1) Abs(x) Log((x + 1)^2/Abs(x)^3))/((x + 1)^2 - Abs(x)^3);

NIntegrate(Integrand(x), {x, -10000000, 10000000})
NIntegrate(Integrand(x), {x, -Infinity, Infinity})
``````

The first instance of `NIntegrate` gives $$2.48398$$, and one can check by plotting the result of `NIntegrate` as a function of the integration limits that the integral appears to converge to this answer for very large integration limits. However, the second instance of `NIntegrate` gives a completely different answer of $$1.75434 times 10^{8}$$. What’s going on here, and how can I make the two integrals agree? I think something that `NIntegrate` might be having trouble with is that the convergence of the integral requires the fact that the integrand is essentially odd for very large $$|x|$$ and these contributions tend to cancel out. Ideally I’d like to find some way to get Mathematica to deal with this properly with the infinite integration range, so that I don’t have to keep remembering to exchange the infinite integration limits with large finite values every time I need to do an integral like this.

## Point wise and uniform convergent of Fourier series

Let $$f$$ be a Riemann integrable function on $$(0,1)$$. What condition(s) on the sequence $${hat{f}(n)}$$ makes the identity $$f(t)=sum_{-infty}^{infty} hat{f}(n)e^{i2pi nt}$$ ?

p.s $$hat{f}$$ is the Fourier transform of $$f$$.

## real analysis – Order of convergence of a product of two convergent sequences

Let $$a_n$$ be a sequence that converges to $$A$$ with order of $$n^alpha$$, that is $$a_n = A + mathcal{O}(n^alpha)$$ and $$b_n$$ is another sequence that converges to B with ordder of $$n^beta$$; i.e. $$b_n = B + mathcal{O}(n^beta)$$. What is the order of convergence of $$a_n cdot b_n$$?

My intuition tells me that:

$$a_nb_n = AB + mathcal{O}(n^{min(alpha,beta)})$$

but could not formulate the proof. I used the trick:

$$|a_nb_n – a_nB + a_nB – AB| leq |a_n||b_n-B| + |B| |a_n-A| leq K n^{alpha}n^{beta}+ Bn^{alpha}$$

## Laplace transform of bounded variation function extends to entire function. Does it imply the integral of the LT is absolutely convergent

I have the following problem. Assume that $$u$$ is a bounded variation function on $$(0,infty)$$. We know that
$$h(z)=int_{0}^{infty}e^{-zy}u(y)dy$$
is absolutely convergent for $$Re(z)>-1$$ and $$h$$ can be extended to an entire function. Does this imply that $$e^{-zy}u(y)$$ is absolutely integrable on $$(0,infty)$$?

I would appreciate any hint and perhaps some reference to a body of literature on Laplace transforms of signed measure provided you are aware of such.

$$I = oint _{|z|=1}frac{dz}{2 pi i z}int_{0}^{infty} dr dfrac{e^{-tfrac{r^2}{z^2}}r^{2n+1}}{z^2(z-1)}$$
• The singular structure in the $$z$$ integral is coming from $$z= 0$$ and $$z = 1$$ points.
• From the term $$e^{-tfrac{r^2}{z^2}}$$ the integral over $$r$$ is giving divergent contribution for $$tfrac{pi}{4} leq Arg(z) leq tfrac{3pi}{4} ; ; & ; ; tfrac{5 pi}{4} leq Arg(z) leq tfrac{7pi}{4}$$
My question is, can we somehow give some argument like analytic continuation or modify the contour so that we can assign $$I$$ to some convergent value?