Mathematics, to steps, needs detailed steps. Mathematics

I just want to say that this is not a result, there is no fixed formula as a result. We can only say that when n approaches positive infinity, it is divergent.

The result of this is divergent, i.e. when n is infinite, its sum is infinite.

Attestation results reference.

Anyone who has studied advanced mathematics knows that harmonic series s=1+1 2+1 3+......is divergent, as evidenced by the following:

Since ln(1+1 n)<1 n (n=1,2,3,...）

So the first n terms of the harmonic series are satisfied and satisfied.

sn=1+1/2+1/3+…+1/n>ln(1+1)+ln(1+1/2)+ln(1+1/3)+…ln(1+1/n)

ln2+ln(3/2)+ln(4/3)+…ln[(n+1)/n]

ln[2*3/2*4/3*…*n+1)/n]=ln(n+1)

Because. lim sn（n→∞）lim ln(n+1)（n→∞）=+∞

So the limit of the SN does not exist, and the harmonic series diverges.

But the limit s=lim[1+1 2+1 3+....+1 n-ln(n)](n) ) exists, because.

sn=1+1/2+1/3+…+1/n-ln(n)>ln(1+1)+ln(1+1/2)+ln(1+1/3)+…ln (1+1/n)-ln(n)

ln(n+1)-ln(n)=ln(1+1/n)

Because. lim sn（n→∞）lim ln(1+1/n)（n→∞）=0

Thus the SN has a nether.

Whereas. sn-s(n+1)=1+1/2+1/3+…+1/n-ln(n)-[1+1/2+1/3+…+1/(n+1)-ln(n+1)]

ln(n+1)-ln(n)-1/(n+1)=ln(1+1/n)-1/(n+1)>ln(1+1/n)-1/n>0

So the SN is monotonically decreasing. From the monotonic bounded series limit theorem, it can be seen that sn must have a limit, therefore.

s=lim[1+1/2+1/3+…+1 n-ln(n)](n) exists.

So let's take this number , which is called Euler's constant, and his approximation is about, and it is not known whether it is rational or irrational. In calculus, Euler's constant has many applications, such as finding the limits of certain sequences, the sum of certain convergent series, and so on. For example, find lim[1 (n+1)+1 (n+2)+....1 (n+n)](n) can do this:

lim[1/(n+1)+1/(n+2)+…1/(n+n)]（n→∞）=lim[1+1/2+1/3+…+1/(n+n)-ln(n+n)]（n→∞）lim[1+1/2+1/3+…+1/n-ln(n)]（n→∞）lim[ln(n+n)-ln(n)]（n→∞）=γ-γln2=ln2

Hu Fangwu shouted the book as follows.

Solution: 1. Direct leveling method.

1）x^2＝25，x＝±5

2) 3x 2 27, x 2 9, elimination x 3

3) (x 5) 2 16, x 5 4, talk with x 5 4, x1 1, x2 9.

4）8（3－x）^2－72＝0，8（x－3）^2＝72，（x－3）^2＝9，x－3＝±3，x1＝6，x2＝0。

5）4x^2-5＝59，4x^2＝59＋5，x^2＝16，x＝±4。

2. Matching method.

1）x^2-4x＋3＝0，（x^2-4x＋4）－4＋3＝0，（x－2）^2＝1，x－2＝±1，x1＝3，x2＝1。

2）x^2-6x＋5＝0，（x^2-6x＋9）－9＋5＝0，（x－3）^2＝4，x－3＝±2，x1＝5，x2＝－1

3) x 2-2x 15 0, x 2-2x 15, x 2-2x 15, x 2-2x 1 16, (x 1) 2 16, x 1 4, x1 5, Na Shi Missing x2 3

4) x 2 x 2x 8 0, x 2 x 2x 8, x 2 x 8 1, (x 1) 3, x1 2, x2

Suppose the vertices of the parabola y=x2-2x+p (a,b)y=x2-2x+p

x-1)2+p-1

a=1b=p-1

The vertex is on the line y=x-1 of 2/2, and (a,b) is substituted for b=(a-1) 2

So p=1

The original can be rewritten as:

y=a(x+m)2+k

x2-2x+1

x-1)2

<> sorry, the last step is wrong, it should be 4-1=3

The only answer to the jujube is like a friend of the rock: good beam.