real analysis - For what values of $x$ in $(-3,17)$ does the series $sumlimits^{infty}_{n=1}frac{(-1)^n x^n}{n[log (n+1)]^2}$ converge?

For what values of $x$ in the following series, does the series converge?

\begin{align}\sum^{\infty}_{n=1}\dfrac{(-1)^n x^n}{n[\log (n+1)]^2},\;\;-3

MY TRIAL

\begin{align}\lim\limits_{n\to \infty}\left|\dfrac{(-1)^{n+1} x^{n+1}}{(n+1)[\log (n+2)]^2}\cdot\dfrac{n[\log (n+1)]^2}{(-1)^n x^n}\right|&=|x|\lim\limits_{n\to \infty}\left|\dfrac{n}{n+1}\cdot\left[\dfrac{\log (n+1)}{\log (n+2)}\right]^2\right|\\&=|x|\lim\limits_{n\to \infty}\left(\dfrac{n}{n+1}\right)\cdot\lim\limits_{n\to \infty}\left[\dfrac{\log (n+1)}{\log (n+2)}\right]^2\\&=|x|\lim\limits_{n\to \infty}\left(\dfrac{n}{n+1}\right)\cdot\left[\lim\limits_{n\to \infty}\dfrac{\log (n+1)}{\log (n+2)}\right]^2\\&=|x|\left[\lim\limits_{n\to \infty}\dfrac{1}{n+1}\cdot n+2\right]^2\\&=|x|\end{align}
Hence, the series converges absolutely for $|x|<1$ and diverges when $|x|>1$.

When $x=1,$
\begin{align}\sum^{\infty}_{n=1}\dfrac{(-1)^n }{n[\log (n+1)]^2}<\infty\;\;\text{By Alternating series test}\end{align}
When $x=-1,$
\begin{align}\sum^{\infty}_{n=1}\dfrac{1}{n[\log (n+1)]^2}<\infty\;\;\text{By Direct comparison test}\end{align}

Hence, the values of $x$ for which the series converges, is $-1\leq x\leq 1.$

I'm I right? Constructive criticisms will be highly welcome! Thanks!

Answer

You are correct. This is a "variation on the theme".

For $|x|>1$
$$\lim_{n\to +\infty}\dfrac{|-x|^n}{n[\log (n+1)]^2}=+\infty$$
and the series is divergent.

For $|x|\leq 1$, by direct comparison, the series is absolutely convergent
$$\sum^{\infty}_{n=1}\dfrac{|-x|^n}{n[\log (n+1)]^2}\leq \sum^{\infty}_{n=1}\dfrac{1}{n[\log (n+1)]^2}<\infty.$$

integration - Help with the integral $int_{0}^{infty}frac{y^{2}e^{y}}{e^{sy}+e^{-sy}-2}dy$

I want to do the integral :

$$I(s)=\int_{0}^{\infty}\frac{y^{2}e^{y}}{e^{sy}+e^{-sy}-2} \, \mathrm{d}y$$
$s$ being a complex parameter. I tried expanding the dominator of the integrand, but this way we lose the symmetry $I(s)=I(-s)$. I tried converting the integral into a contour integral, but I don't know how to close the contour.

Answer

For the integral
\begin{align}
I(s) = \int_{0}^{\infty} \frac{y^{2} \, e^{y} \, dy}{e^{sy} - 2 + e^{-sy}}
\end{align}
it is evident that when $s \to -s$ the result yields $I(-s) = I(s)$. For the evaluation of the integral consider the following.
By making use of
\begin{align}

\frac{1}{(1-x)^{2}} = \sum_{n=0}^{\infty} (n+1) \, x^{n}
\end{align}
then
\begin{align}
I(s) &= \int_{0}^{\infty} \frac{y^{2} \, e^{y} \, dy}{(e^{sy/2} - e^{-sy/2})^{2}} \\
&= \int_{0}^{\infty} \frac{y^{2} \, e^{-(s-1)y} \, dy}{(1 - e^{-sy})^{2}} \\
&= \sum_{n=0}^{\infty} (n+1) \, \int_{0}^{\infty} e^{-(sn + s-1)y} \, y^{2} \, dy \\
&= \sum_{n=0}^{\infty} \frac{n+1}{(sn + s -1)^{3}} \, \int_{0}^{\infty} e^{-u} \, u^{2} \, du \\
&= \frac{2}{s^{3}} \, \sum_{n=0}^{\infty} \frac{n+1}{(n+p)^{3}} \hspace{10mm} p = 1 - \frac{1}{s} \\
&= \frac{2}{s^{3}} \left[ \sum_{n=0}^{\infty} \frac{1}{(n+p)^{2}} + \frac{1}{s} \sum_{n=0}^{\infty} \frac{1}{(n+p)^{3}} \right] \\

&= \frac{1}{s^{4}} \left[ 2s \, \psi^{(1)}\left(1 - \frac{1}{s}\right) - \psi^{(2)}\left(1 - \frac{1}{s}\right) \right]
\end{align}
where $\psi^{(m)}(x)$ are $m^{th}$ derivative of the digamma function.

real analysis - Proof of Raabe's Test for divergence

Question: Prove that if $u_n>0$ and $\lim_\limits{n\to\infty}n\left(\frac{u_n}{u_{n+1}}-1\right)=l$ , then $\sum_{n=1}^{\infty} u_n$ diverges if $l<1$.

My Attempt: We choose a positive number $\epsilon$ such that $l+\epsilon<1$.

Since $\lim_\limits{n\to\infty}n\left(\frac{u_n}{u_{n+1}}-1\right)=l$, then for any $\epsilon>0$ , there exists a natural number $m$ such that $$\left| \,\ n\left(\frac{u_n}{u_{n+1}}-1\right)-l\,\ \right|< \epsilon \,\ \forall \,\ n\ge m$$

Therefore, we have
$$l-\epsilonNow say $$l+\epsilon=r$$ Hence $$r<1$$ So we can say that $$n\left(\frac{u_n}{u_{n+1}}-1\right)

Any suggestions on how to proceed from here?

P.S. I have read other questions here on MathsSE regarding this proof specifically but I am not satisfied with these answers as one uses big O notation and another has a different form of Raabe's test. I don't want answers with big O notation. Proofs with other methods are welcome but the ones following from where I have ended are most preferred.

Answer

Your first steps are all correct.
You have obtained the inequality
$$
\frac{u_n}{u_{n+1}} < \frac{r}{n} + 1
$$
for $n\geq m$
but have realized that, in this form, it is rather hard to do anything with it.

Here is the classical argument as to how to proceed.

Rewrite this as
$$
n \frac{u_n}{u_{n+1}} -(n+1) < r-1 \leq 0.
$$
Then observe that
$$ nu_n \leq (n+1)u_{n+1} $$
for all $n$ greater than $m$. Thus we have an increasing
sequence $\{nu_n\}$ and, in particular there is some positive constant $c$
for which $nu_n\geq c$.

Thus for $n\geq m$ we have $$u_n \geq \frac{c}{n}$$
and a comparison with the harmonic series establishes
divergence as you desired.

There is a more general version of Rabbe's test known as
Kummer's test with very much the same proof.
For that you assume there is a sequence of positive numbers
$D_n$ and you compute the limit
$$
L =\lim_{n\to\infty} \left[ D_n\frac{u_n}{u_{n+1}} -D_{n+1} \right]

$$

Raabe's test is just Kummer's test with $D_n=n$.

The divergence part of Kummer's test (and hence also Raabe's test) doesn't actually require limits. You simply need that $$ D_n\frac{u_n}{u_{n+1}} -D_{n+1} \leq 0$$
for all sufficiently large $n$ and the divergence of $\sum_{n=1}^\infty 1/D_n$ and you can conclude divergence.

measure theory - If $mu(A)

Let $(X,\Sigma,\mu)$ be a finite measure space. Let $\mathcal{F}$ be a family of measurable functions $f:X\to\mathbb{R}$. Prove that if $$\lim_{t\to\infty}\left(\sup_{f\in\mathcal{F}}\int_{\{x\in X:|f(x)|\ge t\}}|f|d\mu \right)=0,$$

then $$\sup_{f\in\mathcal{F}}\int_X|f|d\mu<\infty,$$

and for all $\epsilon >0$ there exists $\delta >0$ such that: $$A\in\Sigma,\mu(A)<\delta\Longrightarrow \sup_{f\in\mathcal{F}}\int_A|f|d\mu<\epsilon.$$

For the first part.

Let $t>0$ such that: $\displaystyle\sup_{f\in\mathcal{F}}\int_{\{x\in X:|f(x)|\ge t\}}|f|d\mu<1$.

Fix $f\in\mathcal{F}$. Then $$\displaystyle\int_X|f|d\mu=\int_{\{|f|\ge t\}}|f|d\mu+\int_{\{|f|

And $1+t\mu(X)$ does not depend of $f$, so we get $\sup_{f\in\mathcal{F}}\int_X|f|d\mu<\infty$.

Is that correct?

I don't know how to do the second part. Could it be true that $v(A):=\sup_{f\in\mathcal{F}}\int_A|f|d\mu$ is a finite measure? I wanted to try something similar to that known result when $\mathcal{F}$ is just one function (some call it absolutely continuous of the measure $v$, I think).

Any hint? Thank you.

Answer

Your first part is correct.

For the second part, try to bound $\int_A |f|\,d\mu$ similarly to how you bounded $\int_X |f|\,d\mu$ in the first part:

$$\int_A |f|\,d\mu=\int_{A\cap \{|f|Then try to make the right side as small as possible, by choosing $t$ and $\delta$ appropriately.

Discontinuous derivative *not* by oscillation

All the differentiable functions that I have ever seen whose derivative is discontinuous, achieve this discontinuity by oscillating: See, e.g., this question.

Is it possible to construct differentiable functions where the discontinuity of the derivative is achieved by some other way?

(The basic idea achieving discontinuous behavior by osciallation is: Let it oscillate around the point $x_0$ where we want our derivative to be discontinuous, but decrease the amplitude as we approach $x_0$ so that we can achieve differentiability. Then the derivative will oscillate as well, but will not descrease in amplitude, so will be discontinuous at $x_0$.)

Answer

For a function of one variable (from $\mathbf{R}$ to $\mathbf{R}$), the derivative (if defined everywhere) can't have a jump discontinuity. This is a consequence of Darboux's theorem, which says that derivatives have the intermediate value property, so that the image of an interval is an interval.

proof writing - Prove $1^3 + 2^3 + cdots + n^3 = (1 + 2 + cdots + n)^2$ using Induction

Prove $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ using Induction

My proof so far:

Let $P(n)$ be $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$

---

Base Case

$P(1):$

LHS = $1^3 = 1$

RHS = $(1)^2 = 1$

Since LHS = RHS, therefore base case holds

---

Induction Hypothesis

Let $n \in \mathbb{N}$ be arbitrary

Assume $P(n)$ holds

---

Induction Step

Prove $P(n+1)$ holds:

$$
\begin{align}
& 1^3 + 2^3 + \cdots + \;n^3 + (n+1)^3 \\
= {} & (1 + 2 + \cdots + \; n)^2 + (n+1)^3 \text{ (by Induction Hypothesis)} \\
= {} & (1 + 2 + \cdots + \; n)^2 + (n^3 + 3n^2 + 3n + 1)
\end{align}
$$

---

This is where I get stuck. I don't know how to prove that my last step is equivalent to:

$$(1 + 2 + \cdots + \;n + (n+1))^2$$

Answer

So basically you want that last expression to turn out to be $\big(1+2+\ldots+n+(n+1)\big)^2$, so you want $(n+1)^3$ to be equal to the difference

$$\big(1+2+\ldots+n+(n+1)\big)^2-(1+2+\ldots+n)^2\;.$$

That’s a difference of two squares, so you can factor it as

$$(n+1)\Big(2(1+2+\ldots+n)+(n+1)\Big)\;.\tag{1}$$

To show that $(1)$ is just a fancy way of writing $(n+1)^3$, you need to show that

$$2(1+2+\ldots+n)+(n+1)=(n+1)^2\;.$$

Do you know a simpler expression for $1+2+\ldots+n$?

Work forward from this...

Okay so going forward.

Basically, Induction works like this, I'll use your question as an example:

Consider the case when $n = 1$. If so, then we will have $1^3 = 1^2$.

Now suppose that $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for some $n \in \mathbb N$.

Recall first that $\displaystyle (1 + 2 + 3 + \cdots + n) = \frac{n(n+1)}{2}$ so we know $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2$.

Now consider $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 + (n + 1)^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2 + (n+1)^3 = \frac{n^2 (n+1)^2 + 4(n+1)^3}{4} = \bigg( \frac{(n+1)(n+2)}{2} \bigg)^2$.

Hence, the statement holds for the $n + 1$ case. Thus by the mathematical induction $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for each $n \in \mathbb N$.

$\mathbb Q.\mathbb E.\mathbb D$

elementary number theory - Finding values for Chinese Remainder Theorem

So I have looked over a lot of the other Chinese Remainder Theorems on here and I still can not completely understand how to answer my question. The question is "Use the construction in the proof of Chinese remainder theorem to find all solutions to the system of congruences."
\begin{align}
x &\equiv 1 \pmod{3} \\
x &\equiv 0 \pmod{4} \\
x &\equiv 1 \pmod{5}
\end{align}

I found my $M=60$, $M_1= 20$, $M_2=15$, $M_3=12$, $a_1=1$, $a_2=0$, $a_3=1$, but I do not understand how to calculate $y_1$, $y_2$, and $y_3$. I think I am supposed to do something with the Euclidian algorithm but I am not sure.