1) a n = 2^ (- n)
2) a n = 100/(n^2)
3) a n = (-1^n)/n
4) a n = sin ( n )/(squareoot(n))
5) a n = 1/(log base10 ( n + 1 ))
In each case, do the following.
(a) Find N ∈ ℕ such that ∀ n ∈ ℕ (n ≥ N ⇒ modulus(a n) %26lt; 1/ 100) . (Note that you do not have to give the best value N , but you must prove all assertions you make.)
(b) Do the same for 1 1000 ...
(c) ... and for 1 10000 .
The following define seqences ( a n ) which all converge to 0.?
11 hours and no replies...I will do |a_n| %26lt; p...you can then put any p tolerance you desire....1) 1 / [2^n] %26lt; p means [1/p] %26lt; 2^n or ln [1/p] %26lt; n ln2...or n %26gt; ln[1/p] / ln 2
2) 100/[n^2] %26lt; p implies [100 / p] %26lt; n^2 or n %26gt; 10 / sqrt p
3) |a_n| = 1 / n...easy one n %26gt; 1 / p
4) |a_n| %26lt; 1 / sqrt n %26lt; p implies 1 / p %26lt; sqrt n or n %26gt; [1 / p^2]
5) 1 / log [n+1] %26lt; p implies 1 / p %26lt; log[n+1] or e^[1 / p ] %26lt; n+1 or n %26gt; e^[1/p] - 1...now b%26amp;c are easy to do {for extra credit?}
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