Advanced Mathematics Review Basics: Sequence Limits

Definition solve

Mainly $e-N$ language

如证${\lim }_{n\to \infty }{q}^n=0$ , where$q$ is a constant and$|q|<1$

untie:

According to the definition, the starting hand is $|a_n-A|<e$

Determine $|q^n-0|=|q^n|<\epsilon$ , using the logarithm for power considerations

Denote $n\ln |q|<\ln e$

Because the left side is negative, take $e\in (0,1)$ , such that$\ln \epsilon$ is also negative

ln > ln ⁡ ε ln ⁡ ∣ q ∣ n>\frac{\ln \varepsilon}{\ln |q| }$n>\frac{\ln }{\ln |q|}$

取$N=[\frac{\ln }{\ln |q|}]+1$

Then when $n>N$ must have$n>\frac{\ln }{\ln |q|}$

则$|q^n-0|<\epsilon$ ,if${\lim }_{n\to \infty }{q}^n=0$ certified

Properties of Convergent Sequences

• (Uniqueness) The existence of a limit must be unique
• (Boundedness) The sequence of limit existence must be bounded
• (Number-guaranteed) sequence { an } \{a_n\}{ an​} limit$a>0$（或$a<0$ ), then there is a positive integer$N$，当$n>N$ has$a_n>0$（或$a_n<0$）

Unequal relationship

Inequality relations mainly rely on inequality (zooming)

Prove that if ${\lim }_{n\to \infty }$, ${\lim }_{n\to \infty }|a_n|=|A|$

untie:

由题意有$\forall \epsilon >0,\exists N>0,n>N,\text{there is}|a_n-A|<e$

Inequality $||a|-|b||\le |a-b|$ get

$$||a_n|-|A||\le |a_n-A|<e$$

Sequence Convergence and Subsequence Convergence

young sequence { an } \{a_n\}{ an​} converges then its sub-column$\left\{a_n\right\}$ also converges, and we have:

$$\underset{k\to \infty }{\lim }{a}_{kn}=\underset{n\to \infty }{\lim }{a}_n$$

This can be said in reverse: if there is a sub-column { an } \{a_n\}{ an​} diverge or the two sub-sequences converge but the convergence values ​​are different, then the original sequence diverges

In some cases, it is necessary to study the sub-sequences to form the whole parent sequence. If the sub-sequence is 2k, then there must be two: 2k and 2k+1. If it is 3k, there must be three: 3k, 3k+1, 3k+2

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Prove that { n ( − 1 ) n } \{n^{(-1)^n}\}{ n(−1)n }limit does not exist

untie:

see $In\; both\; cases\; where\; n$ is odd or even, the convergence limits for odd and even cases are different, so the original number sequence has no limit

other

I don't know what to title

$$\underset{u\to 1}{\lim }{u}^v\stackrel{1^{\infty }}{=}{e}^{\lim v\ln u}=e^{v(u-1)}$$

The premise of establishment is that the form of the formula is $1^{\infty }$ , then obviously$u\to 1$

The origin of this formula is mainly because $\ln u=\ln (1+u-1)$ , and at$x\to When\; 0$ , according to the equivalent infinitesimal,$x\sim \ln (1+x)$ . reason$u\to 1$ , hence$u-1\to 0$ , the above substitution can be applied, and$\ln u=\ln (1+u-1)=x-1$

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Conclusion: When $q$ isa constantand$|q|<1$时，$S={\lim }_{n\to \infty }{S}_n=\frac{a_1}{1-q}$

Emphasis is constant, because if not, see the next question:

When $q=1-\frac{1}{n}$When ${\lim }_{n\to \infty }{q}^{What\; is\; the\; limit\; of\; n}$ ?

untie:

can find that $n\to \infty$ meets$1^{\infty }$ of the form (see conclusion above)

$$\underset{n\to \infty }{\lim }{q}^n=\underset{n\to \infty }{\lim }=e^{{\lim }_{n\to \infty }n(-\frac{1}{n})}=e^{-1}$$

It is not equal to 0 at all

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$$\underset{n\to \infty }{\lim }{a}_n=0\iff \underset{n\to \infty }{\lim }|a_n|=0$$

In this case, to prove $a_n\to 0$ can be transformed into proof$|a_n|\to 0$ , if you use the clamping theorem at this time, you already know$0\le |a_n|\le ...$ , half of the proof is completed, just prove the right side again

The derivation relies on the following inequality:

$$||a_n|-A|$$

On the Pinching Theorem

形式是$$y_n\le x_n\le z_n$$

For two inequality signs, it doesn't matter if there is an equal

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Recursive Prioritization Monotone Bounded Criterion

Set the number sequence { an } \{a_n\}{ an​} satisfya$a_1=a(a>0),a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})$ , prove the existence of the limit and evaluate

untie:

The proof has a lower bound:

Using the inequality $a+b>2\sqrt{ab}$Substitute the right side of the recursive formula to get $a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})=\sqrt{2}$, which is the lower bound

Note that the premise of the inequality is $a,b>0$ , and according to$a_1=a>0$ infers$a_n>0$ , for the premise of the inequality

Certificate monotone:

$a_{n+1}-a_n=\frac{1}{2}(a_n+\frac{2}{a_n})-a_n=\frac{2-a_n^2}{2a{\_}_n}\le 0$ , so monotonically decreasing

Finally solve:

Just bring in A to solve the deduction formula: $A=\frac{1}{2}(A+\frac{2}{A})$ , combined with the guaranteed number$A=\sqrt{2}$

unbounded but not infinite