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

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 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...

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

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.

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)

Prove that every bounded sequence has a convergent
subsequence.

Define a sequence (xn)n≥1 recursively by x1 = 1 and
xn = 1 + 1 /(xn−1) for n > 0. Prove that limn→∞ xn = x exists
and find its value.

1.- Prove the following:
a.- Apply the definition of convergent sequence, Ratio Test or
Squeeze Theorem to prove that a given sequence converges.
b.- Use the Divergence Criterion for Sub-sequences to prove that
a given sequence does not converge.
Subject: Real Analysis

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