Let one of {an} and {BN} be a convergent sequence and the other a divergent sequence. It is proved that {an ± BN} is a divergent sequence We also ask whether {anbn} and {an / BN} (BN ≠ 0} must be divergent sequences

Let one of {an} and {BN} be a convergent sequence and the other a divergent sequence. It is proved that {an ± BN} is a divergent sequence We also ask whether {anbn} and {an / BN} (BN ≠ 0} must be divergent sequences


If {an + BN} converges
Because {an} also converges
For any e
All have N1, N2
If k > N1, there is | (AK + BK) - L | N2, there is | (AK) - a | N1. If the larger of N2 is | (BK - (L-A) | = | (AK + BK) - L + (ak-a) | < | (AK + BK) - L | + | (AK) - a | a = 0 or {BN} - > infinity, then it will be convergent, otherwise it will diverge



How to find this limit? [1 + 3 + 5 +... + (2n-1)] / [2 + 4 + 6 +... + 2n], n tends to be positive infinity
Is there any formula for the limit of sequence and the differential quadrature of numerator and denominator


The top and bottom are equal difference series, which are calculated by the sum of the first n terms of equal difference respectively, and then the coefficient ratio of the highest order term is calculated by division
You can also use a calculator. Well, let's see how close it is



(9N ^ 2 + n) / (2n + 5) find the limit to infinity


Divide up and down by n
=(9n+1)/(2+5/n)
If n tends to infinity, then 5 / n tends to zero
So the numerator tends to infinity and the denominator tends to 2 + 0 = 2
So fractions tend to infinity
So there is no limit



It is known that the equal ratio sequence {BN} is a common ratio Q and the sequence {an} satisfies BN = 3 ^ an. (1) it is proved that the sequence {an} is an equal difference sequence. (2) if B8 = 3 and the sequence {an}
It is known that the equal ratio sequence {BN} is a common ratio Q and the sequence {an} satisfies BN = 3 ^ an, (1) it is proved that the sequence {an} is an equal difference sequence, (2) if B8 = 3 and the first three terms of the sequence {an} are S3 = 39, find the general term of {an}, (3) under the condition of (2), find TN = | A1 | + | A2 | +... + | an|


1.bn/b(n-1)=3[an-a(n-1)]=q
So an-a (n-1) = log (3) Q
2.a2=13
a8=1
d=-2
an=17-2n
3.n8 Tn=-[a1+.an]+2[a1+.+a8
=n^2-16n+128