Question

CONVERGENCE of an INFINITE SERIES

Answer 1

sometimes

Answer 2

@satellite73 here's the question, I don't know where to start proving this problem.

Answer 3

yeah me neither

Answer 4

probably start with the definition of \(\frac{a_n}{b_n}\to L\)

Answer 5

after that i am not sure
this is not a quick one i don't think

Answer 6

what is N?

Answer 7

A large enough number, I suppose.

Answer 8

Example:
\(\sum an=\sum (1/3)^n =1/2 \)
\(\sum bn=\sum (1/2)^n = 1\)
\(Lim_1^\inf (2/3)^n= 0\)
but the quotient is 1/2 (not zero)
However, if N is large enough, say 100, the quotient is 1.8*10^-18...

Answer 9

for small values of n, it converges to L, how can I prove that it is the same even for large values of N? @Callisto can you help me?

Answer 10

I would use a proof similar to a delta-epsilon proof
This problem is really telling you to prove limit comparison test.

So let's say for some epsilon (E) there is an integer n0 so that n > n0
So recall from delta-epsilon proofs, that you have |f(x) - L| < E
So we have |a(n)/b(n) - L| < E ...

Answer 11

So from that, I can expand the abs value to be -E < a(n)/b(n) - L < E
And then add L to both sides to get L - E < a(n)/b(n) < L + E

Answer 12

So usually a(n) is the series we are testing. So b(n) is the comparison series.
So I can get a(n) alone in the inequality by multiply b(n) to all sides.
So (L - E)b(n) < a(n) < (L + E)b(n)

Then we analyze both sides of the inequality separately
So on the left, we have (L - E)b(n) < a(n)
That can also be b(n) < a(n)/(L - E)

Answer 13

So from direct comparison test if (one series) < (2nd series) and the (2nd series) converges that means the 1st series has to converge
So actually on the b(n) < a(n)/(L - E), in order for (series ak) / (series bk) to go to the limit, we need a(n) to be the comparison series and converge
(Sorry I'm not able to draw that out at this time) - glitch ...

So the main point for the left side with b(n) < a(n)/(L - E), in order to get the limit, we need a(n) to converge and then b(n) can converge by direct comparison test and limit comparison would be true

Then on the right side you have a(n) /b(n) < (L+E)
Or a(n) < (L + E)b(n)
Or a(n) / (L + E) < b(n)
So on the right side, b(n) is the comparison series and a(n) is the test series of interest.

So if b(n) converges, then a(n) can converge by direct comparison test there since a(n) is smaller than b(n)

So to do the proof, it's using part of delta-epsilon definition, limit comparison, and direct comparison test.

Answer 14

@darkprince14
What the summations mean is that it may not converge to L for small values of N (the beginning of summation), but for large enough N, it will converge to L.
It is the purpose of my example.