For any ε > 0, there exists n, when n > N, | an + 1-an|

For any ε > 0, there exists n, when n > N, | an + 1-an|

Existing

The general term formula of known sequence an is an = (sin15 ° - cos15 °) ^ n. judge whether the sequence has limit and estimate which constant the limit is

Because 0

Given that the general term formula of sequence {an} is an = [(- 1) ^ n * n] / (2n-1), judge whether the number has limit

The nth power of (- 1) * [n / (2n-1)]
When n tends to infinity [n / (2n-1)] = 1 / 2, but the nth power of (- 1) can not determine whether it is 1 or - 1, so the sequence has no limit

It is known that the sequence whose general term formula is an = (n ^ 2 + a) / N is monotone increasing sequence, then the value range of constant a is__
Condition classification step a

{an} is an increasing sequence, that is, a (n + 1) - an > 0
∵an=(n^2+a)/n=n+a/n
∴a(n+1)=(n+1)+a/(n+1)
∴a(n+1)-an=(n+1)+a/(n+1)-(n+a/n)
=1-A / (n ^ 2 + n) > 0 always holds
For any n ∈ n *, a / (n ^ 2 + n)

The general term formula of sequence an is an = (n-1) / (2n + 3), and its limit is obtained

If the numerator and denominator are divided by N, then
an=(1-1/n)/(2+3/n)
When n →∞, 1 / N → 0, 3 / N → 0,
So an → 1 / 2

The general formulas of sequence {an}, {BN} are an = a * n + 2, BN = b * n + 1 (a, B are constants), and a > B
Then the number of items with the same number and value in the two sequences is
A. 0 B, 1 C, 2 D, infinitely many

Let the kth term of two sequences be equal, then AK + 2 = BK + 1
So 1 = (B-A) k, so k = 1 / (B-A)
And a > b, so b-a

It is known that the common ratio Q of the equal ratio sequence an is a real number, and the sum of the first n terms is Sn, and A3 = 4, S6 = 9s3, so as to find the general term formula of the sequence an

S6=a1(1-q^6)/(1-q)
S3=a1(1-q^3)/(1-q)
S6/S3=(1-q^6)/(1-q^3)=9
(1-q^3)(1+q^3)/(1-q^3)=9
1+q^3=9
q=2
a3=4
a1=a3/q^2=1
an=2^(n-1)

Deriving the sum of the first n terms of an from the equal ratio sequence

The derivation is as follows: because an = a1q ^ (n-1), Sn = a1 + A1 * q ^ 1 +... + A1 * q ^ (n-1) (1) QSn = A1. Q ^ 1 + a1q ^ 2 +... + A1. Q ^ n (2) (1) - (2) note that the first term of formula (1) remains unchanged, and subtract (2) from the second term of formula (1)

It is known that the common ratio Q of the equal ratio sequence an is a real number 1. The sum of the first n terms is Sn and A3 = 4 S6 = 9s3. Find the general term formula of the sequence an 2. Find the first term TN of the sequence n times an

S6 = 9s3, q = 2. A3 = 4, A0 = 1  an = 2 ^ (n-1)
Tn＝∑[1≤k≤n]k2^（k-1）.
See TX = ∑ [1 ≤ K ≤ n] KX ^ (k-1)
There is ∫ [0, x] ttdt = x + X & sup2; + X & sup3; + +x^n＝（x^（n+1）-x）/（x-1）. （x≠1）
Tx＝d｛∫[0,x]Ttdt｝/dx＝d｛（x^（n+1）-x）/（x-1）｝/dx
＝｛nx^（n+1）-（n+1）x^n+1｝/（x-1）²
∴Tn＝Tx|（x=2）＝n2^（n+1）-（n+1）2^n+1.
(verification: T3 = 1 + 2 * 2 + 3 * 2 & sup2; = 17 = 3 * 16-4 * 8 + 1. The formula is correct)

Let the sum of the first n terms of the equal ratio sequence {an} of each real number satisfy S3 = 6, S6 = - 42 to find the general term formula of the sequence {an}?

Let the first term of the sequence {an} be A1 and the common ratio be Q;
S3=A1(1-q³)/(1-q)=6,S6=A1(1-q^6)/(1-q)=-42；
S6/S1=1+q³=-7,∴q=-2；
A1=6(1-q)/(1-q³)=6*(1+2)/(1+8)=2；
General formula: an = 2 * (- 2) ^ (n-1) = - (- 2) ^ n;