## Actual analysis: How to prove that $left ( frac {(- 1) ^ n} { sqrt {n}} right) _ {n geq 1}$ is a null sequence?

How to prove that $$left ( frac {(- 1) ^ n} { sqrt {n}} right) _ {n geq 1}$$ Is it a null sequence?

My attempt for induction:

If I prove that the denominator grows faster than the numerator, I can conclude that it is a null sequence, right? So I have to prove that $$sqrt {n} geq (-1) ^ n$$ for $$forall n in mathbb {N}: n geq 1$$

Base case with $$n_0 = 1$$: $$sqrt {1} = 1 geq (-1) ^ 1 = -1 checkmark$$

Induction hypothesis: $$exists n in mathbb {N}: sqrt {n} geq (-1) ^ n$$

Induction claim: $$Longrightarrow sqrt {n + 1} geq (-1) ^ {n + 1}$$

Inductive step: $$begin {gather} sqrt {n + 1} geq (-1) ^ {n + 1} | cdot (-1) \ – sqrt {n + 1} leq (-1) ^ {n + 2} end {meets}$$ This is true, since $$– sqrt {n + 1}$$ can not be $$geq -1$$.

Is that a valid test?

## Example of $sum a_n$ convergent but $sum sqrt {a_ {2 ^ k}}$ divergent?

The problem is this:

Show that the convergence of $$sum a_n$$ implies the convergence of $$sum frac { sqrt {a_n}} {n},$$ Yes $$a_n geq 0.$$

I am using the Cauchy method:

$$sum frac { sqrt {a_n}} {n}$$ converge $$iff$$ $$sum2 ^ k frac { sqrt {a_ {2 ^ k}}} {2 ^ k} = sum sqrt {a_ {2 ^ k}}$$ converge

Yes $$sum a_n$$ converge $$implies$$ $$sum sqrt {a_ {2 ^ k}}$$ converge is true, I'm finished.

As $$sum a_n$$ converge $$implies$$ $$a_n a 0$$, at least we have $$sqrt {a_ {2 ^ k}} a 0$$. But the thing is, close. $$0$$, squarerooting makes it bigger. So I can not use the comparison test.

But we are choosing $$a_ {2 ^ k}$$ With great distance between them, it will decay much faster. I like it $$2 ^ k$$ Faster. So I'm pretty sure it will converge. However, I can not find $$N$$ such that $$n> N$$ it implies $$sqrt {a_ {2 ^ n}} leq a_n.$$ Actually, I think there would be a counterexample (so the title is).

How do I proceed from here? And is there a counterexample?

## calculation: help is needed with this integral $int frac {6x ^ {3} + 7x ^ 2-12x + 1} { sqrt {x ^ 2 + 4x + 6}} dx$

$$int frac {6x ^ {3} + 7x ^ 2-12x + 1} { sqrt {x ^ 2 + 4x + 6}} dx$$

I received a response from Achille Hui from this thread: How do you evaluate this integral $int frac {6x ^ {3} + 7x ^ 2-12x + 1} { sqrt {x ^ 2 + 4x + 6}} dx$

There are still some points that I do not understand from your answer.

He says set $$B (y) = ay ^ 2 + for + c$$ , RHS (* 1) becomes $$(6-3a) and ^ 3 + …$$ Pin up $$a$$ to $$2$$
$$B (y) = 2y ^ 2 + for + c$$ , RHS (* 1) becomes $$(- 2b-29) and ^ 2 + …$$ Pin up $$b$$ to $$frac {-29} {2}$$

Repeat this procedure once again, we find that c must be set to 24.

I do not understand how he fixes these numbers? Do you leave them equal 0 to get the final result?

## calculation – Can anyone evaluate this indefinite integral for me? Is [integral] e ^ x / sqrt (e ^ x + e ^ (2x) dx

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## precalculus algebra: show the inequality $sum x + 6 ge 2 ( sum sqrt {xy})$

Leave $$x; Y; z in R ^ +$$ such that $$x + y + z + 2 = xyz$$. Such that $$x + y + z + 6 ge 2 ( sqrt {xy} + sqrt {yz} + sqrt {xz})$$

This inequality is not homogeneous and look at the condition that I thought
I would replace the variables $$x; Y; z$$ such that

+)$$x ^ 2 + y ^ 2 + z ^ 2 + 2xyz = 1$$. Leave $$x = frac {2a} { sqrt { left (a + b right) left (a + c right)}}$$

+)$$xy + yz + xz + xyz = 4$$. Leave $$a = frac {2 sqrt {xy}} { sqrt { left (y + z right) left (x + z right)}}$$.Leave $$x = frac {2a} {b + c}$$

but failed. Please explain to me how I can obtain this substitution (if I have a solution by substitution)

I also tried to solve it by $$u, v, w$$.Leave $$sum_ {cyc} x = 3u; sum_ {cyc} xy; Pi_ {cyc} a = w ^ 3 (3u + 2 = w ^ 3; u, v, w> 0)$$ so $$u le w ^ 3-3u$$ or $$4u le w ^ 3$$ but stagnant (I'm very bad at $$uvw$$)

## Simplifying an expression that involves a squareroot: $sqrt {36 – 4x ^ 2}$

I know that $$2 sqrt {9-x ^ 2}$$ is the alternative way of $$sqrt {36 – 4x ^ 2}$$. I tried but I did not realize how to get there. Can someone help?

## Is this enough to prove that sqrt (n) is irrational if n is not a perfect square?

If n is a natural number, then n is a unique product of prime numbers to whole powers

If n is a perfect square, then all the prime factors will be equal to the powers, therefore, when taking the square root, the prime factors of the results will be the whole powers, so the square root is an integer.

If n is not a perfect square, then at least one prime factor is an odd power. Therefore, its square root will have a prime to a non-integer power, but all integers can be written as a product of primes to an integer power, therefore, sqrt (n) can not be an integer.

Is this enough to prove that all square roots of non-perfect squares are irrational?

## nt.number theory – A new formula for the class number of the quadratic field $mathbb Q ( sqrt {(- 1) ^ {(p-1) / 2} p})$?

I have the following conjecture that implies a possible new formula for the class number of the quadratic field $$mathbb Q ( sqrt {(- 1) ^ {(p-1) / 2} p})$$ with $$p$$ an odd cousin

Guess. Leave $$p$$ be an odd cousin and leave $$p ^ * = (- 1) ^ {(p-1) / 2} p$$. Then the class number $$h (p ^ *)$$ of the quadratic field $$mathbb Q ( sqrt {p ^ *})$$ matches the number
$$frac {( frac {-2} p)} {2 ^ {(p-3) / 2} p ^ {(p-5) / 4}} det left[cotpifrac{jk}pright]_ {1 le j, k le (p-1) / 2},$$
where $$( frac { cdot} p)$$ It is the symbol of Legendre.

This is Conjecture 5.1 in my prepress arXiv: 1901.04837. I've checked it for all odd prime numbers $$p <29$$. Note that $$h (p ^ *) = 1$$ for each odd cousin $$p <23$$Y $$h (-23) = 3$$.

Here I invite some of you to review this conjecture more thoroughly. My computer can not verify it $$p = 29$$.

## reference request – The Iwasawa $lambda$ variable of the cyclotomic $mathbb {Z} _3$ – extension of $mathbb {Q} ( sqrt {-3})$

Now I am in an investigation about the Iwasawa $$lambda$$-Invariances of the cyclotomy. $$mathbb {Z} _p$$-extensions of numerical fields. And it happens that the cyclotomics. $$mathbb {Z} _3$$-extension of $$mathbb {Q} ( sqrt {-3})$$ It is an elementary but important case for my initial investigation.

But I could not find a literature on it yet, and I'm not an expert in this area either, so I'm not used to computing invariants alone.

So please can you give any information about this $$lambda$$-invariant?

Thank you.

## real analysis: find $x$ such that $5 ^ x – sqrt {2x} – log_2 {x} = 22$

Find $$x$$ such that $$5 ^ x – sqrt {2x} – log_2 {x} = 22$$.

I have observed that the solution to this equation should be $$x = 2$$, I also graficé the graph of the function $$f (x) = 5 ^ x – sqrt {2x} – log_2 {x}$$, and it seems a growing function, so the solution must be unique.

However, my problem is to prove that this solution is unique and that the function is increasing using its derivative does not seem to work, since the expression is quite ugly.

Do you have any suggestions on how to solve this without using the derivative?