Find the sum of series n 1 1 n 1n 2n 1 3n

The result: <>

The process of solving the problem is as follows:

Series n 1 (?1)n?1n(2n?1) The sum of 3n and the series n 1 xln(3?x)，x∈[-3，3]．

Solution s(1 3) 3 9 ?2ln2+ln3．：

Introduce the power series n 1 (?1)n?1 n(2n?1) x2n （.x . 1）

Therefore let s(x) = n 1 (?.)1)n?1 n(2n?1) x2n, then s(1 3 )=

n＝1 ?1)n?1

n(2n?1)3n

s″（x）=22

n＝1 (?1)n?1x2n?1＝2

n＝1 ?x2)n?1＝2 1+x2 （.x .＜1）

Combined with s(0)=0, s(0)=0, so s(x)=2arctanx, then integral has.

s（x）=2

x0 arctantdt=2[arctanx-∫ x0 t 1+t2 dt]

2xarctanx-ln（1+t2）| x0

2xarctanx-ln（1+x2）（x2＜1）．

Therefore the series n 1 xln(3?x)，x∈[-3，3]．

Introduce power series

n＝1n?1n(2n?1)x2n

x, so s(x)=

n＝1n?1n(2n?1)x2n

then s(1> n 1

n?1n(2n?1)n

s″x）=22∞

n＝1n?1x2n?1

n＝1?xn?11+x

x conjugates s(0)=0,s

0)=0, so s(x)=2arctanx, and then integrals have.

s（x）=2∫x

arctantdt

2[arctanx-∫x

t1+tdt2xarctanx-ln（1+t2

x2xarctanx-ln（1+x2（x2

Therefore, s(12LN2+LN3

We consider n to be from 0 to infinity.

Solution: Let f(x) = ((1) n (2n+1))x (2n+1), then: f'(x)= ((x 2) n=1 (1+x 2), (x|<1)

So: f(x)=arctanx, when x= 1, the series is staggered and still converges, so f(x)=arctanx, (|x|《1)

Let x=1 yield: (1) n (2n+1)=f(1)=arctan1= 4

(-1)^n/(2n+1)=(-1)^n*(1)^(2n+1)/(2n+1)

Let s(x) = (-1) n*x (2n+1) (2n+1)s'(x)=(∑(1)^n*x^(2n+1)/(2n+1))'=∑[(1)^n*x^(2n+1)/(2n+1)]'=∑(-1)^n*x^2n=(-x^2)^n

1 (1+x 2) (sum of proportional sequences).

So s(x) = 1 (1+x 2)dx=arctanx, so (-1) n (2n+1)=s(1)=arctan1= 4, I count n=0 to n=

1 (2n 1x1)-1 (3nx1).

n [(2n dec. 1) (3n dec. 1)].

So. 1 [(2n ten 1) (3n ten 1)].

1 n (2n dec. 1) - 1 n (3n dec. 1).

2 [1 2n-1 (2n dec. 1)]-3 [1 3n-1 (3n dec. 1)]1 n-2 (2n dec. 1)-1 n dec. 3 (3n dec. 1)3 (3n dec. 1)-2 (2n ten-1).

Since (n>=1)[(2n-1) (2 n)] n>=1)[n2 (n-1)].

n>=1)[(1 2) n], while.

n>=1)[(1 2) n], for seeking. (n>=1)[n/2^(n-1)]

, do power series.

f(x)(n>=1)[n(x 2) (n-1)], using the item-by-item integration theorem, g(x) is obtained

0,x]f(t)dt

n>=1)n [0,x][(t 2) (n-1)]dt2 (n>=1)[(x 2) n], derivative, get.

f(x).

f(x)…, from which it can be obtained.

n>=1)[n/2^(n-1)]

1. The solution to this question is:

a. Seek guidance first; Then, b, then take advantage of the common ratio less than 1

the formula for summing the proportional sequence of infinite terms; Finally, c. re-integrate, and get the answer.

2. The idea of solving the problem in this way is:

An infinite series, within the convergence domain, is strictly equal to a function, and this function is the sum function.

Because they are strictly equal, so, at the same time, the derivative does not affect their equivalence;

At the same time, definite integrals do not affect their identity.

With the above thoughts, it is logical to seek all derivatives and then integrate, and then integrate them.

3. The specific answers are as follows:

∑（∞n→0）(2n+1)x^n

r=lim|2n-1/2n+1|=1

( n 0)2n+1) diverges when x=1, and ( n 0)(-1) n(2n+1) also diverges when x=-1, so the convergence domain is (-1,1).

Let s(x) = ( n 0)(2n+1)x n= ( n 1)2nx n+ (n 0)x n

Re-order ( n 1)2nx n=s1(x).

s1(x)=2x∑（∞n→1）nx^(n-1)

2x∑（∞n→1）(x^n)'

2x(∑（n→1）x^n)'

2x[x/(1-x)]'

2x/(1-x)^2

and ( n 0)x n=1 (1-x).

So s(x)=2x(1-x) 2+1 (1-x)=(1+x) (1-x) 2

, n 0)(2n+1)x n=(1+x) (1-x) 2, x belongs to (-1,1).

-1) n (2n+1) = (1) n*(1) (2n+1) (2n+1) let s(x) = 1) n*x (2n+1) (2n+1)s'(x)=(1)^n*x^(2n+1)/(2n+1))'1)^n*x^(2n+1)/(2n+1)]'1) n*x 2n=(-x 2) n=1 (1+x 2) (sum of proportional sequences) so s(x) = 1 (1+x ..).

Let s(x)= x (2n)] 2n+1) multiply both sides xxs(x) = x (2n+1)] 2n+1) is the derivative of the time.

xs(x)) = x 2n = 1 (1-x 2) and then the two divide and jerk back.

xs(x) 1 (1-x 2)ln[(1+x) branch with hunger (1-x)].

s(x)=(1 2x)ln[(1+x) (1-x)] x is not 01 x 0 let x=(root number 2) 2 substituting s(x) is what is s(x) is s.