I'm trying to show that limn→∞xn=limn→∞1√n|n∑k=1(−1)k√k|=12.
Assuming limn→∞xn=x exists, we have
x2n=√2n−1(−x2n−1)+√2n√2n
Letting n→∞,
x=−x+1
x=12
But I'm stuck on proving the existence of limxn. Any idea?
Update: I just solved the problem using sandwich theorem + integral test. Still, I would like to see a continuation of my initial idea, i.e. proving
{x2n} is monotonically increasing (similarly, {x2n+1} is monotonically decreasing)
Answer
limn→∞1√n|n∑k=1(−1)k√k|=limn→∞1√2nn∑k=11√2k−1+√2k
12√2nn∑k=11√2k<1√2nn∑k=11√2k−1+√2k<12√2nn∑k=11√2k−1<12√2n(1+n−1∑k=11√2k)
With Riemann we get:
1√2nn∑k=11√2k=12nn∑k=11√k/n→121∫0dx√x=√x|10=1
And therefore the confirmation of the claim.
About the monotonie.
sn:=n∑k=11√2k−1+√2k
Assumption about the monotonie: sn√2n is strictly increasing.
sn√2n<sn+1√2n+2 leads to
(1√2n−1√2n+2)sn<1√2n+21√2n+1+√2n+2 and therefore to
sn<an:=√2n2√2n+2+√2n√2n+2+√2n+1
For n=1 it’s o.k. . Assume it’s correct for n . (∗)
Then we have to show that it's also correct for n→n+1 .
sn+1<√2n2√2n+2+√2n√2n+2+√2n+1+1√2n+1+√2n+2<an+1
The first inequation follows from the assumption (∗) and the second inequation can be proved by some transformations, 2n replaced by x and x≥0:
(√x(√x+2+√x)+2)(√x+4+√x+3)<
<√x+2(√x+4+√x+2)(√x+2+√x+1)
Transformations:
(√x(√x+2+√x)+2)(√x+4+√x+2)+(√x(√x+2+√x)+2)(√x+3−√x+2)<
<√x+2(√x+4+√x+2)(√x+2+√x+1)
(√x(√x+2+√x)+2)(√x+3−√x+2)<(√x+2(√x+2+√x+1)−(√x(√x+2+√x)+2))(√x+4+√x+2)
(√x√x+2+x+2)(√x+3−√x+2)<(√x+2√x+4+x+2)(√x+1−√x)
This is true because of:
√x√x+2+x+2<√x+2√x+4+x+2
√x+3−√x+2<√x+1−√x
Note: √x+1+a−√x+a is decreasing by growing a>−x
No comments:
Post a Comment