Limits of a Sequence Homework: Find and Prove Answer

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SUMMARY

The discussion focuses on finding the limits of sequences defined by the greatest integer function, specifically for a_n = [1/n] and a_n = [(10+n)/(2n)]. The first sequence converges to zero as n approaches infinity, confirmed by the condition 1/n < ε. The second sequence also converges to zero, but the analysis reveals that the floor function accelerates convergence rather than slowing it down, as shown by the expression (n+10)/(2n) simplifying to 1/2 + 5/n.

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Homework Statement


Let [x] be the greatest integer ≤x. For example [itex][\pi ]=3[/itex]
and [3]=3
Find [itex]lim a_n[/itex] and prove it.
a) [itex]a_n=[\frac{1}{n}][/itex]
b) [itex]a_n=[\frac{10+n}{2n}][/itex]

The Attempt at a Solution


for the first one it will converge to zero.
so can I write [itex]\frac{1}{n}< \epsilon[/itex]
then I can just pick an n large enough to make that work.
for part b, it also looks like it will converge to zero, but a little slower.
so [itex]\frac{10+n}{2n}< \epsilon[/itex] and then solve for n in terms of ε.
Or is there something I am not taking into account with the greatest integer deal?
 
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for the first one:

you are NOT trying to "solve for ε in terms of n", but rather, the other way around. you need to find a (possibly large) value for N so that n > N makes 1/n "small" (less than epsilon). epsilon is arbitrary, but assumed as given.

for the second one, i would note that:

(n+10)/2n = 1/2 + 5/n.

if you find an N such that n > N means 5/n < 1/2,

wouldn't [(n+10)/2n] = 0 for all such n?

it seems to me the floor function "speeds up" the convergence, instead of making it slower.
 
thanks for your help, ok
so on the second one [itex].5+\frac{5}{n} < \epsilon[/itex]
so then I just pick an n large enough to make it less than epsilon.
 

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