You can use the comparison test. Since the convergent sequence divided by n is less that the convergent sequence, it must converge.
How would you prove algebraically that the following function is one to one? f(x)= (x+3)^2 , x>= -3?
Looking at the graph of the function can give you a good idea. However, to actually prove that it is even or odd may be more complicated. Using the definition of "even" and "odd", for an even function, you have to prove that f(x) = f(-x) for all values of "x"; and for an odd function, you have to prove that f(x) = -f(-x) for all values of "x".
The concept of contrapositive comes from here. A implies B is equivalent to not B implies not A. Prove by contrapositive means instead of proving condition A leads to B, show that if B fails also cause A to fail.
A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular
(x-y)2 is a square so (x-y)2 >= 0 expanding, x2 - 2xy + y2 >= 0 so x2 + y2 >= 2xy or 2xy <= x2 + y2
The limits on an as n goes to infinity is aThen for some epsilon greater than 0, chose N such that for n>Nwe have |an-a| < epsilon.Now if m and n are > N we have |an-am|=|(am -a)-(an -a)|< or= |am -an | which is < or equal to 2 epsilor so the sequence is Cauchy.
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.
Cannot prove that 2 divided by 10 equals 2 because it is not true.
A logical sequence in an argument is a way to prove a step has a logical consequence. Every proposition in an argument must be tested in this fashion to prove that every action has a reaction.
Let (an) be a convergent sequence in metric space X. Let "a" be the limit of an. Then for sufficiently large N, if x,y>N, d(ay,a)<e/2 and d(ax,a)<e/2 for any e greater than zero. We add these inequalities and get d(ay,a) + d(ax,a)<e. But by the triangle inequality d(ay,ax)<d(ay,a) + d(ax,a), so d(ay,ax)<e for all x,y>N. (some of those < should be less than or equal to, but it doesn't really affect the proof.)
It has not yet been proven whether any arbitrary sequence of digits appears somewhere in the decimal expansion of pi.
We prove that if an increasing sequence {an} is bounded above, then it is convergent and the limit is the sup {an }Now we use the least upper bound property of real numbers to say that sup {an } exists and we call it something, say S. We can say this because sup {an } is not empty and by our assumption is it bounded above so it has a LUB.Now for all natural numbers N we look at aN such that for all E, or epsilon greater than 0, we have aN > S-epsilon. This must be true, because if it were not the that number would be an upper bound which contradicts that S is the least upper bound.Now since {an} is increasing for all n greater than N we have |S-an|
It is REM(4n/6)=4. Prove. Correction.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. 6 is an even number.
ask yourself if anything can go into 57 or 111 (divided)
Because there is no way to define the divisors, the equations cannot be evaluated.
Long division 8/11 = 8 divided by 11 .7272727272recurring