how to determine whether xn=(a^n+b^n)^(1/n) convergent or divergent

Question

anonymous · Answer

$$x_{n}=(a_{n}+b_{n})^{1/n}$$

experimentx · Answer

what are an and bn??

anonymous · Answer

oh sory...$$x_{n}=(a^{n}+b^{n})^{1/n}$$....0<a<b

experimentx · Answer

$ a( 1 + (a/b)^n)^{1/n}$
= $ 1 + 1/n (a/b)^n + (1/n)(1/n-1)/2! (a/b)^{2n}+ (1/n)(1/n-1)(1/n -3)/3! (a/b)^{3n}$ <--- let's forget a for a while

experimentx · Answer

sorry ... (1/n - 3) was supposed to be (1/n - 2)

experimentx · Answer

as n->inf (a/b)^n ->0 (confused at the moment)
so i guess it will converge to 
a

anonymous · Answer

how you get a(1+$$(a/b)^{n})^{1/n}$$

experimentx · Answer

take a^n common

anonymous · Answer

but how cn get (a/b)[^{n}]?not (b/a)[^{n}]?

experimentx · Answer

?? i am not understanding ... latex is not rendering.

experimentx · Answer

looks like I made mistake somewhere.

anonymous · Answer

$$why (a/b)^{n}?not  (b/a)^{n}?$$

experimentx · Answer

oo .. yeah ... that's my mistake.

experimentx · Answer

let's take b^n out of there ... and let's see what happens.

experimentx · Answer

looks it converges to b http://www.wolframalpha.com/input/?i=%282%5E100+%2B+3%5E100%29%5E%281%2F100%29

anonymous · Answer

this one to find limit..how to show it is convergent?

experimentx · Answer

show that it converges to b as  n-> infinity.

experimentx · Answer

the same process as above, but instead of a, we take out b. expand it binomially, and then take limit n->inf, we must get result b.

anonymous · Answer

i thnk we hve to show first it is cnvrgent..thn we fnd the limit..mybe cn use any test..

experimentx · Answer

is it related to real analysis??

experimentx · Answer

http://www.sosmath.com/calculus/sequence/morelim/morelim.html http://www.tricki.org/article/To_prove_that_a_sequence_converges_find_one_or_more_convergent_subsequences

anonymous · Answer

yup

experimentx · Answer

we can prove that is is bounded.

anonymous · Answer

can you prove it is cnvergent and bounded?

experimentx · Answer

a^n + b^n < (a+b)^n

it seems that it is always bounded by a+b, which implies it is convergent.

anonymous · Answer

ok2..we hve to show it is decreasing,then bounded.by monotonic cnvergence theorem,xn is cnvergence...then find the limit

anonymous · Answer

is there any theorem said that if the set bounded then it converge???

anonymous · Answer

theorem only said a cnvergent sequence is bounded.

anonymous · Answer

yup...it doesn't mean that bounded sequence is convergent

experimentx · Answer

if an increasing sequence is bounded above, then it is convergent
similarly, if decreasing sequece is bounded below, then it is also converget.
it seems the sequence is clearly bounded above and below (b-a)

anonymous · Answer

i got it..thanks..

experimentx · Answer

you are welcome.