Proof 1 1 2 ... 1 n is an infinite amount

1+1/2+..1/n

1 + 1/2 + 1/3 + 1/4) +1/5 + 1/6 + 1/7 + 1/8) +1/9 + 1/10 + 1/16)+ 1/n

1 + 1/2 + 1/4 + 1/4) +1/8 + 1/8 + 1/8 + 1/8) +1/16 + 1/16) +1/n

1 + 1 2 + 1 2 + 1 2 + 1 n (when n -> infinitely large).

1 + n/2

When n -> infinitely massive, n2 -> infinitely massive, 1 + n2 -> infinitely mass.

1+1/2+..1 n > infinitely large, so 1+1 2+...1 n is an infinite amount

This harmonic series, the harmonic series is the case of p series p=1, and the basic common sense is that the harmonic series diverges, that is, infinity, this is a theorem. As for how to prove it. This can be proved here:

Let s=1+1 2+. 1/n

1+1/n】+【1/2+1/（n-1）】+1/m+1/（n-m+1）】

n+1）/n+（n+1）/【2（n-1）】+n+1）/【m（n-m+1）】

If the denominator is the same, let's take the reciprocal.

1 s can be calculated as tending to o, then s tends to infinity.

It's time to get off work, and it seems to be wrong in my thinking, so I'll think about it when I go back

Summary. Hello and happy to answer your questions.

Original = lim(1+2+......n)/n^2

lim[n(n+1)/2]/n^2

1/2lim(n+1)/n

1/2*lim(1+1/n)

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How to solve this problem.

Hello, it's a pleasure to answer your questions=lim(1+2+......n) n 2=lim[n(n+1) 2] n 2=1 2lim(n+1) n=1 2*lim(1+1 n)=1 2 2*1=1 2 I hope my help to search for your relatives, if you have any questions, please feel free to ask me <>

Can you write down the detailed process?

n is how it was solved.

The primordial hail is transformed into a percolation sail limn(n+1) 2n 2=lim(n+1) 2n=1 2 The original is lim(x+1) [root number(x 2+x)+root number(x 2-1)]=lim(1+1 x) [root number(1+1 x)+root number(1-1 x)]=1 2

s(n)=1 Chongjube 1+1 2+1 scattered 3+.There is no limit to the number 1 n. That is, this series is divergent, not convergent.

The following proves that s(n) can reach infinity:

1 5 + 1 6 + 1 7 + 1 8 >=1 8)*4 >=1 2

So: (2 n is the nth power of 2).

s(2^n)>=1/2)*n+1.

So there is no limit to s(n)!

Solving high-math problems, infinitesimal proofs. When n, the nth power of n2 is infinitesimal.

un=n 2 n, obviously when n tends to infinity, n and 2 n tend to infinity, so to meet the conditions used by Lopida's law, the dust fiber can be derived from the numerator and denominator at the same time, that is, when n tends to infinity, lim n 2 n=lim (n).'2^n)'=lim 2n ln2*(2 n), obviously n and 2 n still tend to infinity, so continue to derive the numerator denominator lim 2n ln2*(2 n)=lim 2 ln2) 2 n), at this time, the denominator tends to infinity, and the numerator is a constant collapse, so the polar group brother dismantles the limit value tends to 0, that is, infinitesimal.

1 Companion 2+1 4+1 8+1 16+1 32+....+1 2 n1-1 coarse Kai 2 n

When the rock calls n infinite, 1 2 n = 0

So infinite series 1 2 + 1 4 + 1 8 + 1 16 + 1 32+....The sum of is 1

1/3>1/4

So 1 3+1 4>1 4+1 4=1 21 5+1 6+1 7+1 8>1 8+1 8+1 8+1 8=1 2

In the same way, 1 9 + 1 10 + ......1/16>8*1/16=1/21/17+……1 32>16*1 32=1 2, so 1+1 2+......1/n>1+1/2+1/2+1/2+……Here there are infinitely many 1 2 added to the right, so it is infinity f(x)=1 (x-1) 2 is the infinite mass when x 1 and f(n)=n 2 is the infinite quantity when n. The reciprocal of an infinitely large quantity is an infinitesimal quantity. It should be especially noted that constants, no matter how large, are not infinitely large.

There are different definitions of infinity in set theory. The German mathematician Cantor proposed that the number of elements (cardinality) corresponding to different infinite sets has different "infinity". The sum of two infinitely large quantities is not necessarily infinity, the product of a bounded quantity and an infinitely large quantity is not necessarily infinity (as in the case of a constant 0 is a bounded function), and the product of a finite infinite quantity must be infinite.

1/3>1/4

So 1 3+1 4>1 4+1 4=1 21 5+1 6+1 7+1 8>1 8+1 8+1 8+1 8=1 2

In the same way, 1 9 + 1 10 + ......1/16>8*1/16=1/21/17+……1/32>16*1/32=1/2……So 1+1 2+......1/n>1+1/2+1/2+1/2+……Here there is an infinite number of 1 2s on the right that are added so it is infinity so the left side is infinity.

Partial sum = 1 + 1 2 + 1 3) +1 4 + 1 5 + 1 6 + 1 7) +

1 + 2*1 4 + 4*1 8 + 1 + 1 2 + 1 2 + 1 2 + so it is an infinite amount .

The 1 (n +1) after you draw the line is convergent because 1 (n +1) < 1 n

According to the p-series theory, 1 n is convergent.

It can be seen that the primary series 1 (n +1) is convergent.

But the first n terms before the key are: 1= n

It's divergent, so there's no way to determine its convergence.

This problem should be dealt with with the convergence point requisite, if liman ≠ 0, then the series an must diverge!

Obviously lim n (n +1) = 1≠0, so the series is divergent!

Your questions are as follows:

The first thing you're wrong about is that parentheses are omitted; It should be written:

Sigma (1 - 1 (n 2 + 1)) Sigma 1 - Sigma (1 (n 2 +1)) and then talk about the result:

Indeed, the latter term, when n—> =1, but the sum of the former term =n;

Thus, sigma(n2(n2+1)) n-1;