# Thread: help with sequences

1. ## help with sequences

Prove:
lim (KSn) = KS and lim (K + Sn) = K + S, for any k element of R

Sn means S sub n

2. Originally Posted by slowcurv99
Prove:
lim (KSn) = KS and lim (K + Sn) = K + S, for any k element of R

Sn means S sub n
So first we need to define what it means to be a limit. There ar etwo major definitions, which are essentially the same. I have used both, the first from Calc 1 to calc 3 and the latter from the proofs class i'm doing now. I will prove the first question using both methods, so you can see how they are similar and different, and the last by the latter definition only. Here are the definitions:

Definition 1:

Let be a function defined on an open interval containing (except possibly at ) and let be a real number. means that
for each real there exists a real such that for all where we have .
Definition 2:

We say the limit of a sequence (Sn) is a real number, s, provided that for each e > 0 there exists a natural number N such that n>N implies |Sn - S| < e

Note: e means epsilon. d means delta

lim (KSn) = KS
Proof (By definition 1): The result is obvious for k = 0. Assume k is not 0. Let limSn = S. So for each e>0 there exists a d>0 such that 0<|x - c|<d implies |Sn - S|<e/|k|.

Then 0<|x - c|<d implies |kSn - kS| <or= |k||Sn - S| < |k|.e/|k| = e

So we have for each e>0, there is a d>0 such that 0<|x - c|<d implies |kSn - kS|< e. Thus the limkSn = kS

QED

Alternate Proof (By definition 2): The result is obvious for k = 0, therefore, let us assume k is not 0.

Let e>0. Since limSn = S, there exists an N such that n>N implies |Sn - S|< e/|k|

So n>N implies |kSn - kS| <or= |k||Sn - S| < |k|.e/|k| = e

Since n>N implies |kSn - kS| < e, limkSn = kS.

QED

lim (K + Sn) = K + S
Proof (By Definition 2): The result is obvious for k = 0, so let's assume k is not 0. Let the limSn = S. This means for any e > 0, there exists an N such that n > N implies |Sn - S| < e

Then n>N implies |(K + Sn) - (K + S)| = |Sn - S + K - K| = |Sn - S| < e

Since n>N implies |(K + Sn) - (K + S)| < e, lim(K + Sn) = K + S

QED

The proof by definition 1 is very similar as you can see in the first question I did. You should be able to reproduce this result using that definition if that's the one you learned in class.

3. Originally Posted by Jhevon
So first we need to define what it means to be a limit. There ar etwo major definitions, which are essentially the same. I have used both, the first from Calc 1 to calc 3 and the latter from the proofs class i'm doing now. I will prove the first question using both methods, so you can see how they are similar and different, and the last by the latter definition only. Here are the definitions:
.

These are not limits of functions, it is limits of sequences.

lim (KSn) = KS
Let me doth this one, and leave ye other to thee.

Let {s_n} be a convergent sequence.
Define lim s_n =s.

There are two cases to consider.

Case 1, k=0
There is nothing to prove because k*s_n=0 which is a konstant sequence.
Q.E.D.

Case 2, k !=0
For any e>0 we have, e/|k|>0.
Thus, there exists an N such that,
|s_n - s|<e/|k|
Multiply by |k| to get,
|s_n-s|*|k|=|ks_n-ks|<e
Thus, {ks_n} is convergent and lim k*s_n=ks.

Q.E.D.