# Answer on Real Analysis Question for steve peters

Question #16571

assume that lim[1+2(-1)^n]Xn = 0. Prove that lim Xn exists and find it.

Expert's answer

if n=2k then lim[1+2(-1)^n]Xn=lim[1+2(-1)^(2k)]X2k=lim3*X_(2k)

if n=2k+1 then

lim[1+2(-1)^(2k+1)]Xn=lim(-1)*X_(2k+1)

then for subsequences

0=lim3*X_(2k)=3lim X_(2k) and 0=lim(-1)*X_(2k+1)=-lim X_(2k+1)

so we

have lim X_(2k)=lim X_(2k+1)=0 so limit exist and equal 0

if n=2k+1 then

lim[1+2(-1)^(2k+1)]Xn=lim(-1)*X_(2k+1)

then for subsequences

0=lim3*X_(2k)=3lim X_(2k) and 0=lim(-1)*X_(2k+1)=-lim X_(2k+1)

so we

have lim X_(2k)=lim X_(2k+1)=0 so limit exist and equal 0

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment