By S. Zaidman.

Ch. 1. Numbers --
ch. 2. Sequences of genuine numbers --
ch. three. limitless numerical sequence --
ch. four. non-stop capabilities --
ch. five. Derivatives --
ch. 6. Convex capabilities --
ch. 7. Metric areas --
ch. eight. Integration.

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Extra info for Advanced calculus : an introduction to mathematical analysis

Example text

Then bn — an > 0 for n > n 0 , and A — B > 0. We also have B — e < bn < B + e for n > ni, A — e < an < A + e for n > n 2 . Therefore bn — an < B + £ — (A — e) = B — A + 2e for n > max (ni, n 2 ), also 0 < frn — an < B — A 4- 2e for n > max(n 0 , ni, n 2 ). On the other hand, B - A < 0 and B - A + 2e < 0 for 2e = ^ - . We obtain 0 < 0, a contradiction. Next, we shall prove the following: Theorem 5. Let (a n ) be a convergent sequence, such that a n > OVn G N and lim an = a. T/ien; VA; G N, we have lim a£ = a*=.

Example 2. Consider the sequence: xn — (1 4- ; ^ ) 3 T T ' • We write rrn = [(l+^2)n+2'(l + ^2)-2]3. Then (z ) = ((1 + ; r b ) n + 2 ) i s a subsequence of ((1 4- ^ ) m ) , and accordingly lim zn n — e. Obviously lim (1 4- r r o ) 2 = 1. ). Example 3. We examine the sequence (xn) where xn — (1 4- J ) n . )*-H)"(^)*-K)"(^r(^rFrom this relation we easily find that L = lim (1 + J ) n = e • e • 1 = e 2 as guessed previously. Example 4. Let us define a sequence (an) in the following way: ai = \/2, a 2 = V 2 + \/2, a 3 = Y 2 + V 2 + v ^ , .

O Let us assume, conversely, that ^ ^ ( 1 + ^ ) * is convergent. This entails- lim ak(l +ak)~^ =0 ° Therefore, 3k~ e N, such that, for k > k~, ofc(l + a f c ) - 1 < ±. Then we find 2ak < 1 + ak and a^ < 1 for k > k . Therefore, we see that again for k > k~ we have < o flfc (in fact, if afc = 0, we have nothing to prove. If ak > 0, we see that 1 < Y ^ «*• 1 + o* < 2 o ofc < 1, hence afc < ^ ) . OO oo From these reasonings we derive that £ > * is convergent. The two series are simultaneously convergent, hence simultaneously divergent.

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