Question

Given that xn is a sequence of real numbers. If (xn) is a convergent sequence prove that (xn) is bounded. That is, show that there exists C > 0 such that |xn| less than or equal to C for all n in N.

Answer #1

If (x_n) is a convergent sequence prove that (x_n) is bounded.
That is, show that there exists C>0 such that abs(x_n) is less
than or equal to C for all n in naturals

Prove that if (xn) is a sequence of real numbers,
then lim sup|xn| = 0 as n approaches infinity. if and
only if the limit of (xN) exists and xn
approaches 0.

If (xn) ∞ to n=1 is a convergent sequence with limn→∞ xn = 0
prove that
lim n→∞ (x1 + x2 + · · · + xn)/ n = 0 .

If (xn) is a sequence of nonzero real numbers and if limn→∞ xn =
x where x does not equal zero; prove that lim n→∞ 1/ xn = 1/x

Let
<Xn> be a cauchy sequence of real numbers. Prove that
<Xn> has a limit.

) Let α be a fixed positive real number, α > 0. For a
sequence {xn}, let x1 > √ α, and define x2, x3, x4, · · · by the
following recurrence relation xn+1 = 1 2 xn + α xn (a) Prove that
{xn} decreases monotonically (in other words, xn+1 − xn ≤ 0 for all
n). (b) Prove that {xn} is bounded from below. (Hint: use proof by
induction to show xn > √ α for all...

Prove or disprove that if (xn) is an unbounded sequence in R,
then there exists n0 belongs to N so that xn is greater than 10^7
for all n greater than or equal to n0

Suppose {xn} is a sequence of real numbers that converges to
+infinity, and suppose that {bn} is a sequence of real numbers that
converges. Prove that {xn+bn} converges to +infinity.

Exercise 2.4.5: Suppose that a Cauchy sequence {xn} is such that
for every M ∈ N, there exists a k ≥ M and an n ≥ M such that xk
< 0 and xn > 0. Using simply the definition of a Cauchy
sequence and of a convergent sequence, show that the sequence
converges to 0.

For Xn given by the following, prove the convergence
or divergence of the sequence (Xn) with a formal proof,
clearly and neatly:
a) Xn = n2/(2n2+1)
b) Xn = (-1)n/(n+1)
c) Xn = sin(n)/(n2+1)

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