Question

If (xn) ∞ to n=1 is a convergent sequence with limn→∞ xn = 0 prove that

lim n→∞ (x1 + x2 + · · · + xn)/ n = 0 .

Answer #1

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

Define a sequence (xn)n≥1 recursively by x1 = 1, x2 = 2, and xn
= ((xn−1)+(xn−2))/ 2 for n > 2. Prove that limn→∞ xn = x exists
and find its value.

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.

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.

) 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 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.

7.5.
Prove the following:(a) lim n→∞ an = a = ⇒ lim n→∞ |an|=|a|. (b)
limn→∞an=0 ⇐ ⇒ lim n→∞|an|=0

Let xn be a sequence such that for every m ∈ N, m ≥ 2 the
sequence limn→∞ xmn = L. Prove or provide a counterexample: limn→∞
xn = L.

Let n ≥ 2 and x1, x2, ..., xn > 0 be such that x1 + x2 + · ·
· + xn = 1. Prove that √ x1 + √ x2 + · · · + √ xn /√ n − 1 ≤ x1/ √
1 − x1 + x2/ √ 1 − x2 + · · · + xn/ √ 1 − xn

Let 0 < θ < 1 and let (xn) be a sequence where
|xn+1 − xn| ≤ θn for n
= 1, 2, . . ..
a) Show that for any 1 ≤ n < m one has |xm −
xn| ≤ (θn/ 1-θ )*(1 − θ m−n ).
Conclude that (xn) is Cauchy
b)If lim xn = x* , prove the following error in
approximation (the "error in approximation" is the same as error
estimation in Taylor Theorem) in t:...

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