Number series root number n 2 2 root number n 1 root number n , find the limit of the sum of t

a(n) = [(n+2)^(1/2) -n+1)^(1/2)] n+1)^(1/2) -n^(1/2)],s(n) = a(1)+a(2)+.a(n-1)+a(n)

3^(1/2)-2^(1/2)]-2^(1/2)-1^(1/2)] 4^(1/2)-3^(1/2)]-3^(1/2)-2^(1/2)] n+1)^(1/2)-n^(1/2)]-n^(1/2)-(n-1)^(1/2)] n+2)^(1/2)-(n+1)^(1/2)]-n+1)^(1/2)-n^(1/2)]

n+2) (1 2)-(n+1) (1 2)] 2 (1 2)-1 (1 2)],1 [(n+2) (1 2) +n+1) (1 2)] 2 (1 2) +1,n-> infinity, 1 [(n+2) (1 2)+(n+1) (1 2)] 0, so when n-> infinite, s(n) -0 - 2 (1 2) +1 = 1-2 (1 2).

Solution: Because an= (n+2)*[n+1)- n-1)] n+2)*2 [ (n+1)+ n-1)] The numerator is physicochemical)2 (1+2 n) [ (1+1 n)+ 1-1 n)] The numerator and denominator are divided by n).

So, n time 1 n 0

So an 2 (1+1)=1, i.e. lim(n)an=1.

Solve limn no Ming Tomb imitation poor.

n+1）-√n-1））

limn infinity.

limn no Wang Sun poor (2 ( n+1)+ n-1)))limn infinity 2 (2 n).

limn infinity 1 n

n No excitation of fiber poverty.

The denominator can be rationalized, for example, 1 ( m+ (m+1)) and multiply the numerator and denominator by ( ( (m+1)-m) to obtain 1 ( m+ (m+1))=m+1)-m

Then sum. sn=√(n+1)-1=10

n=120 hope it helps.

First, the original formula is divided into parts, and the fraction is multiplied by n+1 under the root number - n under the root number, then the original formula becomes n under the root number n+1 - n under the root number, that is, the general term of the series is the root number n + 1 - the root number n, so sn=a1+a2++an again because a1 = root number 2 - 1a2 = root number 3 - root number 2a3 = root number 4 - root number n + 1 - root number 2 in root number NA1 and A2.

Root number (1*2) + root number (2*3) +The root number n(n+1)> 1 + 2 + n = n(n+1) 2

Root number (1*2) + root number (2*3) +Root number n(n+1)< n+1) +n+1) +n+1) =n*(n+1).

1 = root number (n+1) - root number n number of the first n term of the series of travel closure = root number (n + 1) - root number 1 = 10 n = 120

This is too simple, by the square difference skin destruction formula [ ( (n+1) +n][ n+1) -n]=(n+1)-n=1, i.e. 1 (root number (n+1) + root number n) = n+1) -n, so there is sn= 2-1+ 3- 2+ number grip 4- 3·· n+1) -n, but the middle of the potato mold is all gone,..

an= (n+2)*[n+1)- n-1)] n+2)*2 Huaiyan does not have jujube to do [ (n+1)+ n-1)] molecule is physicochemical)

2 lead sodium (numerator and denominator divided by (n+2)).

So, when n, an 2 (1+1)=1, i.e., lim(n)an=1