How can I prove that $lim_{n to infty} x_{n} = x Longleftrightarrow x_{n} to x$ and $midmid x_{n} midmid to midmid x midmid$ when $n to infty$?

# Tag: limn

## $lim_{n rightarrow infty} frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1}$. Am I correct?

I have to find this limit:

begin{align} lim_{n rightarrow infty}

frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1} end{align}

**My attempt:**

Since we know that $-1leq sin (x) leq 1$ for all $x in mathbb{R}$ we have:

begin{align}

lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdot(-1)}{n+1}underbrace{leq}_{text{Is this fine?}}&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{(1)^{n}cdotsqrt{n}cdot(1)}{n+1}\ \ lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{n}}{n+1}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{n}}{n+1}

end{align}

My doubt in the inequeality is because of the $(-1)$ terms. If everything is correct, then we have

begin{align}

lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{frac{1}{n}}}{1+frac{1}{n}}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{frac{1}{n}}}{1+frac{1}{n}}\ \ Rightarrow 0 leq &lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq 0 \ \ therefore& lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1}=0

end{align}

Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!

## Problem proving $lim_{n rightarrow infty} int_S(f(x))^{frac{1}{n}}=mu(S)$

Statement:

Given a measurable set $S subseteq mathbb{R}$ and a measurable function $f:S longrightarrow mathbb{R}$ with $f(x)gt1$ for all $x in S$. Show that

$$lim_{n rightarrow infty} int_S(f(x))^{frac{1}{n}}=mu(S)$$

I have searched that there are some similar questions here and here, but they do not start from the same hyphothesis.The sequence $f_n = (f(x))^{frac{1}{n}}$ is decreasing and I have tried to solve it by dominated/monotone convergence, but i believe that there is something missing about the integrability of $f$. I do not know how to continue.

## nt.number theory – Generalization of $lim_{n rightarrow infty} prod_{i=1}^{n}frac{2n-1}{2n}$ for a character $chi:mathbb{Z}/s mathbb{Z} rightarrow mathbb{C}^*$

Playing with some infinite products I came up with this problem, that I’m not able to figure it out by myself. Moreover in the internet it doesn’t seem to appear anywhere.

Maybe it is just an easy consequence of properties of characters that I’m not aware of, anyway thank you in advance for any help/answers/suggestions.

All of us know (it is fairly easy to see) that $$lim_{n rightarrow infty} frac{1 cdot 3cdot dots cdot (2n-1)}{2 cdot 4 cdot dots cdot (2n)}=0$$

Now this fact could be reformulated in this fashion: let $$chi:mathbb{Z}/2 mathbb{Z} rightarrow mathbb{C}^*$$

the only non trivial character of $mathbb{Z}/2 mathbb{Z}$, then the expression $$lim_{n rightarrow infty}(prod_{n in mathbb{N}}n^{chi(n : text{mod} 2mathbb{Z})})^{-1}=0$$

More generally, we can perform this construction for every $s in mathbb{N}$.

Indeed all we have to do is to consider a non trivial character $$chi:mathbb{Z}/s mathbb{Z} rightarrow mathbb{C}^*$$

and consider the limit

$$lim_{n rightarrow infty}(prod_{n in mathbb{N}}n^{chi(n : text{mod} smathbb{Z})})^{-1}$$

Now it is true that:

- The value of the limit is finite for every $s$ and every non trivial character $chi$?
- If so the value of the limit depends only on $s$ or also on $chi$?
- The limit is always a real number? (Possibly $0$?)

## $ lim_{n to +infty} frac{1}{sqrt{n}} | sum_{k=1}^{2n} (-1)^k sqrt{k}| $ Did I do it right? How to do the last step?

I need to find:

$$ lim_{n to +infty} frac{1}{sqrt{n}} | sum_{k=1}^{2n} (-1)^k sqrt{k}| $$

So I calculate:

$$ lim_{n to +infty} frac{1}{sqrt{n}} | sum_{k=1}^{2n} (-1)^k sqrt{k}| = lim_{n to +infty} |frac{sum_{k=1}^{2n} (-1)^k sqrt{k}}{sqrt{n}} | $$

Denominator grows to $infty$, so I can use stolz theorem:

$$lim_{n to +infty} |frac{sum_{k=1}^{2n} (-1)^k sqrt{k}}{sqrt{n}} | = lim_{n to +infty} |frac{sqrt{2n}}{sqrt{n} – sqrt{n-1}} | implies(frac{sqrt{2}}{0} ) $$

So I have to transform that a bit:

$$lim_{n to +infty} |frac{sqrt{2n}(sqrt{n} + sqrt{n-1})}{n – n + 1} | = lim_{n to +infty} |sqrt{2}n + sqrt{2n^2-2n}| implies infty – infty$$

Another transformation:

$$lim_{n to +infty} |frac{2n^2 – 2n^2 + 2n}{sqrt{2}n – sqrt{2n^2-2n}}| = lim_{n to +infty} |frac{2n}{n(sqrt{2} – sqrt{2-frac{2}{n}})}| = lim_{n to +infty} |frac{2}{sqrt{2} – sqrt{2-frac{2}{n}}}| implies(frac{2}{0} )$$

… and I feel like it doesn’t lead nowhere. Can somebody tell what do I do wrong?

## real analysis – Show $lim_{n to infty} int_0^1 g(x^n) dx = g(0)$.

**Assume $g$ is integrable on $(0, 1)$ and continuous at $0$. Show
$lim_{n to infty} int_0^1 g(x^n) dx = g(0)$.**

My attempt at this question: I split up the integral into two parts from $(0,1-alpha)$ and $(1-alpha,1)$. For $(0,1-alpha)$, there is uniform continuity, so the limit $n to infty$ can be shifted inside the integral to get $g(0)$.

But how do I keep the integral over $(1-alpha,1)$ to be small?

Note: This is Exercise $7.4.10$ in Abbott, Understanding Analysis, 2nd edition.

## sequences and series – Find the limit: $lim_{n to +infty}(frac{1^p + 2^p + … + n^p}{n^p} – frac{n}{p + 1})$, where $p in mathbb{N}$

Find the limit: $lim_{n to +infty}(frac{1^p + 2^p + … + n^p}{n^p} – frac{n}{p + 1})$, where $p in mathbb{N}$.

I’ve got an idea to transform this sequence into $lim_{nto +infty}(frac{x_n}{y_n})$ and to use Stolz’s theorem.

But in 2 days I didn’t get ideas for this transformation.

## real analysis: a sequence $ {a_n } $ is such that $ lim_n (a_ {n + 1} – a_n) = 0 $. Given some additional properties of $ a_n $ test that converges.

Leave $ {a_n } $ be a sequence such that:

$$

lim_ {n to infty} (a_ {n + 1} – a_n) = 0 tag1

$$

The sequence also satisfies the following properties:

$$

begin {align *}

| a_ {n + 2} -a_ {n + 1} | le | a_ {n + 1} -a_ {n} | tag 2 \

(a_ {n + 2} – a_ {n + 1}) (a_ {n + 1} -a_n) le 0 tag 3

end {align *}

$$

Try out $ {a_n } $ converge

While looking at the problem for the first time, I was thinking that $ (1) $ it implies convergence of $ {a_n } $, but looking more carefully one can think of a harmonic series:

$$

a_n = sum_ {k = 1} ^ n {1 over k}

$$

Which is clearly divergent as $ n a infty $, even though $ (1) $ still maintains. So obviously there is an extra job to do. So below are some of my additional discoveries. First let's define a new sequence:

$$

b_n = a_n – a_ {n-1} tag 4

$$

It is given that:

$$

0 | b_ {n + 2} | le | b_ {n + 1} | le cdots le | b_1 |

$$

For this $ | b_n | $ it is a monotonically decreasing sequence limited then, therefore, convergent. If now we take a look at $ (3) $ There are two possible cases for the declaration, either:

$$

b_ {n + 2} ge 0 \

b_ {n + 1} le 0 tag5

$$

or:

$$

b_ {n + 2} le 0 \

b_ {n + 1} ge 0 tag6

$$

But at the same time:

$$

b_ {n + 2} b_ {n + 1} le 0 \

b_ {n + 1} b_ {n} le 0 \

b_ {n} b_ {n-1} le 0 \

cdots

$$

Let's stick to the case $ (5) $ for the definition This implies:

$$

0 le b_ {n + 2} le b_n le b_ {n-2} le cdots tag7

$$

At the same time from $ | b_n | $ is decreasing monotonously and $ b_ {n + 1} le 0 $:

$$

cdots le b_ {n-3} le b_ {n-1} le b_ {n + 1} cdots le 0 tag8

$$

It seems that $ b_n $ is an alternating sequence, consisting of two convergent sub-sequences as well as $ 0 $, that is because $ (7) $ is decreasing monotonously towards $ 0 $ Y $ (8) $ is increasing monotonically towards $ 0 $. And this depends on whether we assume $ (5) $ or $ (6) $.

That's where I got stuck. I do not see how to combine those findings to show that $ a_n $ Actually it is convergent. How do I proceed?