Question

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

Answer #1

if you have any doubt regarding any step you can ask in comment i will clarify there

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.

Suppose that an is a sequence and lim sup an =a=lim inf an for
some a in Real numbers.
Prove that an converges to a

let xn be a
sequence in R do lim sup xn and lim inf xn always exist R # and
why

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.

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.

Show that lim sup n→∞ (−xn) = −(lim inf n→∞ (xn).

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

prove if lim?→∞ an = a>0 and if lim?→∞ sup bn = b (bn≥0) then
lim?→∞ sup anbn =ab
0<R<∞ : an∈R

If lim Xn as n->infinity = L and lim Yn as n->infinity =
M, and L<M then there exists N in naturals such that Xn<Yn
for all n>=N

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