# Sum of two limits equals limit of the sum

Discussion in 'Undergraduate Math' started by boris, Aug 27, 2005.

1. ### borisGuest

hi

I keep trying to understand the proof for that the sum of the limits is equal to the limit of the sum but to no effect .

we have two sequnces - a and b

e ... degree of closeness
L(a) ... limit of sequence a
L(b) ... limit of sequence b

If there are such natural numbers N(a) and N(b) , that for each

n > N(a): |a - a(n)| < e/2 and n > N(b): |b - b(n)| < e/2

-First thing that confuses me is why do we pick e/2 for both sequences . Why not 2/3*e for a and 1/3*e for b ?

Continuing with the proof :

If we pick the larger of two Ns , then for each n > N

|(a+b) - ( a(n)+b(n) )| = |a - a(n) + b - b(n)|

|a - a(n) + b - b(n)| < |a-a(n)| + |b-b(n)|

|a-a(n)| + |b-b(n)| < e/2 + e/2 = e

How does above prove anything ? I don't get it

thank you

boris, Aug 27, 2005
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2. ### Mike TerryGuest

equal to the limit of the sum but to no effect .
Why not 2/3*e for a and 1/3*e for b ?
No fundamantal reason. Using e/2 for both sequences seems more natural, but
the proof would still works the way you are suggesting.
Setting c(n) = a(n) + b(n), c = a + b, what you are trying to prove is:

c(n) ---> c.

The conclusion you have reached above (expressed in terms of c(n)) is:

for each n > N
|c - c(n)| < e

which is exactly what you were aiming for...

Regards,
Mike.

Mike Terry, Aug 27, 2005
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3. ### Jim SpriggsGuest

You could do it like this:

-e < a(n) - L(a) < e for all n > N1

-e < b(n) - L(b) < e for all n > N2

is what you're given: the two sequences have their respective limits.

So

-e/2 < a(n) - L(a) < e/2 for all n > max(N1, N2)

-e/2 < b(n) - L(b) < e/2 for all n > max(N1, N2).

Add those to get

-e < a(n)+b(n) - (L(a)+L(b)) < e for all n > max(N1, N2).

Hence

a(n) + b(n) ---> L(a) + L(b).

Jim Spriggs, Aug 27, 2005
4. ### VirgilGuest

What one should be aiming for is:

for every e > 0 there is an n such that m > n implies |c - c(m)| < e

Or, equivalently,

for every e > 0, the set { n : |c - c(n)| >= e} is finite.

Virgil, Aug 27, 2005
5. ### Ken OliverGuest

You could in this problem pick e/3 and 2/3*e, but it is more convenient to
use e/2 both times.

Ken Oliver, Aug 28, 2005
6. ### borisGuest

But if limit for sequence a is 2 and for b is 3 , wouldn't then be more accurate that limit for a+b is 4.99999... ? I'm not shure why , but somehow I imagine limit for a+b would be a little less then 2 + 3 = 5

Shouldn't sign be <= instead of < ? Anyhow , what is the point of above equation?
What did we prove by showing that right side is greater than left side ?

boris, Aug 28, 2005
7. ### Paul SperryGuest

I can't imagine why you would think that and, anyway, 4.9999... isn't
"a little less" than 5. For a discussion of infinite length on this
topic see the NG sci.math.
You are trying to show that if a_n -> a and b_n -> b
then (a_n + b_n) -> a + b.

Skratch work: I need |(a_n + b_n) - (a + b)| < e and I know
that |a_n - a| < e and |b_n - b| < e. Well,
|a_n + b_n - (a + b)| = |(a_n - a) + (b_n - b)| <=
||a_n - a| + |b_n - b| < e + e = 2e. Aha!

Proof: Let e > 0. Apply the definition of sequence convergence and
choose N and M such that if n > N then |a_n - a)| < e/2 and if n > M
then |b_n - b| < e/2. Let K = max(N, M). Then, for all n > K,
|(a_n + b_n) - (a + b)| <= |a_n -a| + |b_n - b| < e/2 + e/2 = e. Thus,
by definition, (a_n + b_n) -> (a + b).

Paul Sperry, Aug 28, 2005
8. ### borisGuest

But if limit for sequence a is 2 and for b is 3 , >>>wouldn't then be more
If we would only have a(n) sequence with limit=2
then a(n)+3=5 and limit=5 would make sense sincefor however close a(n) would be to 2,a(n)+3 would be just as close to 5

But if a(n) + b(n) then a(n)+b(n) is a little less closer to 5 then a(n)+3 and thus was wondering if perhaps the limit cold be number smaller than 5

What does -> sign mean in this context ?

boris, Aug 28, 2005
9. ### Paul SperryGuest

Suppose a_n converges to 3 (a_n -> 3 to answer your question below) and
b_n -> 2. Suppose we have N so that if n > N then a_n is within 0.6 of
3. Then, sure enough, a_n + 2 is within 0.6 of 5 for the same n's. If
we pick M so that if m > M then b_m is within 0.6 of 2 and if we let K
be the larger of N and M and let k > K then it is true that the best we
can say is that a_k + b_k is within 1.2 of 5.

_But_, all we need to do is make K bigger to guarantee that a_k + b_k
is within 0.6 of 5. (Just make K larger than the N and M that get a_n
and b_m within 0.3 of 3 and 2 resp. - hence the "e/2".)

You may have to go "farther out" in the a_n + b_n sequence to get
within 0.6 of 5 than you do in the a_n and b_n sequences to get within
0.6 of 3 and 2 but that's OK.

[...]
"Converges to"

Paul Sperry, Aug 29, 2005
10. ### MJRGuest

But 4.99999... does equal 5.

If you don't absolutely believe that, . . .

(1) can you find for me a number "y" such that

4.99999... < "y" < 5 ?

If you can, then 4.99999... is definitely < 5.
If you cannot, then 4.99999... = 5, case closed.

(2) note that the sum

4 + 9/10 + 9/100 + 9/1000 + 9/10000 + ...

converges to 5. After the "4", let's set up

9/10 + 9/100 + 9/1000 + 9/10000 + ...

as 9 * (1/10 + 1/100 + 1/1000 + 1/10000 + ... )

and let's write it as 9 * (r + r^2 + r^3 + ... ),
where r = 1/10.

The sum r + r^2 + r^3 + ... is r / (1 - r) [I'm happy
to provide a quick proof of that, if it's unfamiliar],
which is going to be (1/10) / (1 - 1/10), which turns
out to be (1/10) / (9/10), which in turn is (1/9).

ASIDE

Of course this relies on the concept of the sum of
an infinite series, so conceptually we can turn out
to be be back where we started!

END OF ASIDE

But then 9 * (r + r^2 + r^3 + ... ) = 9 * (1/9), which
is going to be 1.

Then 4.99999... = 4 + 1, which [by golly] is 5.

M J R

MJR, Aug 29, 2005
11. ### Stan BrownGuest

Plus ca change ...

Stan Brown, Aug 29, 2005

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