If the sequence (n) converges to a limit L then, by definition, for any eps>0 there exists a number N such |n-L|N. However if eps=0.5 then whatever value of N we chose we find that whenever n>max{N,L}+1, |n-L|=n-L>1>eps. Proving the first statement false by contradiction.
Chat with our AI personalities
1/6 n(n+1)(n+2)
Descending (in a sequence) means that a the next number is "more negative" or "closer to negative infinity" or "less positive" or "further from positive infinity" or if n is a number in a sequence and n+1 is the next number then n/n+1 > 1
If x(n) represents the nth number is the sequence x(n+1)=x(n)*(x(n)+1) So the next number in the sequence is 1086*(1086+1)=3263442
The set of odd numbers is an arithmetic sequence. Let say that the sequence has n odd numbers where the first term is a1 and the last one is n. The formula to find the sum on nth terms for an arithmetic sequence is: Sn = (n/2)(a1 + an) or Sn = (n/2)[2a1 + (n - 1)d] where d is the common difference that for odd numbers is 2. Sn = (n/2)(2a1 + 2n - 2)
The sum of a sequence is given by sum = n/2(2a + (n-1)d) where: n = how many a = first number of sequence d = difference between terms of sequence. For the first 22 odd numbers these are: n = 22 a = 1 d = 2 → sum = 22/2(2×1 + (22 - 1)×2)) = 22² = 484 The sum of the first n odd numbers is always n²: sum = n/2(2×1 + (n-1)2) = n/2(1 + (n-1))×2 = n(n) = n²