Find the sum Sn of the first n terms of the sequence 1 n

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

Without a specific formula, this value tends to infinity.

I don't know if you have learned natural logarithms, derivatives, and definite integrals, but if you do, it will prove that it is greater than ln(n+1).

Dear, this is a harmony progression, you can't get it directly, he's divergent, and it's hard to ask for it, and I generally won't call you to ask for peace, right!?

The sum of the first n terms of the series.

When you count the occasional argument, s=-1+2-3+4-......n-1)+n1×(n/2)

n 2 When n is an odd balance, s=-1+2-3+4-......n-1)-n1×(n-1/2)-n

n-1）/2

n+1）/2

Summary. Split term method, which is the specific application of decomposition and combination ideas in the summation of sequences. It is to decompose each term (general term) in the series and then recombine it so that some terms can be eliminated and finally the purpose of summing is achieved.

The relationship between multiples of general term decomposition (split term). It is usually used for algebra, fractions, and sometimes integers.

an 3 n(n+1), find the sum of the first n terms of the series.

This problem uses the method of elimination of split terms.

1 When summing by the method of elimination of split terms, the general term should be transformed, such as: n k(1) k(1) ( nn k(1) k(1)n k(1), after the split term can produce successive terms that cancel each other out. 2 It is not necessary that only the first and last items will be left after offsetting, and it is also possible that there are two items left in the front and two items in the back.

The split-term method is the specific application of decomposition and combinatorial thinking in the summation of sequences. It is to decompose each term in the series of numbers (general terms), and then recombine them so that they can eliminate the base and infiltrate some terms, and finally achieve the purpose of summing. The relationship between multiples of general term decomposition (split term).

It is usually used for algebra, fractions, and sometimes integers.

This is the formula for expansion.

From the formal point of view, it is a general term that combines equal difference and equal ratio, and the term elimination method can be adopted.

an = n+1)*(1 2) (n+1) The common ratio of a proportional series is 1 2

sn = 2*(1/2)^2 + 3*(1/2)^3 + n*(1/2)^n+ (n+1)*(1/2)^(n+1)

1/2sn = 2*(1/2)^3 + n*(1/2)^(n+1)+ n+1)*(1/2)^(n+2)

Subtract the two formulas and find sn

Solution: by algebraic formula 1

nn(n1)(2n

1) 6, then the sum of the first n terms of the series

annn(n1)(2n

To sum up, what the original question asked. sn

n(n1)(2n

1)/6，nn

(1) When n is an even number, let n=2k, then k=n 2sn=1 -2 +3 -4 +....2k-1)²-2k)²=(1-2)(1+2)+(3-4)(3+4)+…2k-1-2k)(2k-1+2k)

1-2-3-4-……2k-1)-2k=-(2k+1)*2k/2

k(2k+1)

n(n+1)/2

2) When n is odd, let n=2k-1, then k=(n+1) 2sn=1 -2 +3 -4 +....2k-3)²-2k-2)²+2k-1)²

1-2)(1+2)+(3-4)(3+4)+…2k-3-2k+2)(2k-3+2k-2)+(2k-1)²

1-2-3-4-……2k-3)-(2k-2)+(2k-1)²=-(2k-1)*(2k-2)/2+(2k-1)²=k(2k-1)

n(n+1)/2

In summary, sn=(-1) (n+1)*n(n+1)2

Use the "dislocation subtraction" summation method.