Help prove an inequality 1 1 2 1 3 1 2 to the nth power 1 n 2

This is proved by mathematical induction:

In the first step, when n=1, 1 > 1 2 holds.

In the second step, if n=k, 1+1 2+1 3+....+1 (2 k-1) >k 2.

Then n=k+1, 1+1 2+1 3+....+1/[2^(k+1)-1]

1+1/2+1/3+…+1/(2^k-1) +

A total of 2k items were noted.

Because 2 k < 2 k+1 < 2 (k+1)-1 < 2 (k+1).

So: >2 k) [1 2 (k+1)]=1 2

So: 1+1 2+1 3+....+1 [2 (k+1)-1] >k 2+1 2=(k+1) 2 holds.

Finally, it is concluded that 1+1 2+1 3+....+1 (2 n-1) >n 2 is established.

In the first step, when n=1, 1 > 1 2 holds.

In the second step, if n=k, 1+1 2+1 3+....+1 (2 k-1) >k 2.

Then n=k+1, 1+1 2+1 3+....+1/[2^(k+1)-1]

1+1/2+1/3+…+1/(2^k-1) +

A total of 2k items were noted.

Because 2 k < 2 k+1 < 2 (k+1)-1 < 2 (k+1).

So: >2 k) [1 2 (k+1)]=1 2

So: 1+1 2+1 3+....+1 [2 (k+1)-1] >k 2+1 2=(k+1) 2 holds.

Finally, it is concluded that 1+1 2+1 3+....+1 (2 n-1) >n 2 is established.

This conclusion is wrong, 1 2 + 1 3 + 1 4 ......+1/n>1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)

.1/2^k+1/2^k+..There are jujube back 2 k 1 2 k) stool slag +.1/2^k)

1/2+1/2+1/2+..1 2 tends to be + the world.

The equation for summation of equations and columns is that when q≠1 sn=a1(1-q) (1-q)=(a1-anq) (1-q).

Q=1 when sn=na1

a1 is the first term, an is the nth term, d is the volt tolerance, q is the isoridge staring ratio) a1=1, an=3 (n-1), q=3, sn=[1-3*3 (n-1)] 1-3))=3 n 1) 3 1)).

Method 1: Use mathematics to summarize the disadvantages and draw Li method: prove it yourself;

Method 2: Yinbei because of: 1 21 (2n+1).

c can be seen that the resulting number square is the first number plus half of the second number 2011 2 4022

So choose C and you observe each given formula, don't care about the square, you will find that the number you get is the number in the middle, and when you look at the middle number, you will find that it is half of the first and last digits, this is a problem to find the law, there is no deeper why.

Proof: Using the second mathematical induction.

Proof of it. 1. When n=1, the proposition is clearly true. Namely:

1/2＜ln3-ln2＜1(1);Let the proposition be true when n k. i.e. when n=2, there is: 1 3 ln3-ln2 1 2(2);.

Place the first k-2 inequalities.

The two sides are added together: 1 2+1 3+1 4+. 1/(k-1)＜ln(k-1)＜1+1/2+1/3+1/4+。。

1 (k-2) (*When n=k-1, there is: 1 k lnk 1 (k-1). Add the left and right ends of the first k-1 inequality to get :

1/2+1/3+1/4+。。1/k＜ln(k-1)＜1+1/2+1/3+1/4+..1 (k-1) (*-(*)

1/k＜lnk-ln(k-1)＜1/(k-1)。So the proposition is true for n=k-1. Let's take a look at what happens when n=k.

2. When n=k, it is necessary to prove: 1 (k+1) ln(k+1)-lnk 1 k. Use the method of counter-evidence.

to prove it. Assuming that the proposition does not hold true for n=k, i.e., 1 (k+1) ln(k+1)-lnk 1 k does not hold, then there is:

ln(k+1 k) 1 k and ln(k+1 k) 1 (k+1) or -ln(k+1 k) -1 (k+1) holds. So the + formula yields: 0 1 k-1 (k+1)=1 k(k+1).

So there is: k(k+1) 0, but k 0. So the solution to this inequality can only be:

k＜-1。This is a positive integer with the laughing sock hole dismantling k.

Contradiction! So the assumption is incorrect. To sum up, the proposition is for all natural numbers.

n are all true.

1/1²+1/2²+1/3²+.1 n <2 when n = 1, 1 on the left and 2 on the right, and the inequality holds when n 2 1 n < 1 [(n-1)n]=1 (n-1)-1 n 1 1 =1

1/n²<1/(n-1)-1/n

Add the two sides: 1 1 +1 2 +1 3 +1 n <2-1 n<2 In summary, there is always for any n n*.

Equation 1 1 +1 2 +1 3 +1/n²<2

Solution: Use the deflation method: multiply the denominator by 1 times 2, 2 times 3, 3 times 4...

n-1 multiplied by n, and we have: 1 (1 2) 1 (2 3) ....1 (n minus 1 times n) 1-half + one-half ) one-third)....n minus one-half) n one-part = 1-n one, proven.

Observe the following equations:

1+3+5+..2n-1)=[2n-1+1)/2]²=n²;

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an=2n-1 is an equal difference series, with the first term being 1 and the tolerance being 2

sn=n(a1+an)/2＝n(1+2n-1)/2=2n^2/2=n^2