P-Series Test: Does 1/An→∞ as n→∞?

[cloud360]

Let An be a sequence. If An==>0 as n==> infinity, then is it necessarily the case that 1/An==>infinity as n==>infinity

I am using mobile version of physis forums:

Attempted solution

i know the p series test says 1/n^p converges if p>1

 

[nonequilibrium]

Are you familiar with epsilon proofs?

 

[cloud360]

hi, i am not trying to prove it. this is true or false questions.

i am familiar with find N(epsilon) and proving something is a limit as it tends to a non infinity number using epsilon and some other symbol

 

[nonequilibrium]

Okay, so I think you'd agree the basic question is "can 1/A_n stay bounded if A_n tends to zero". To find out, you can either to it visually or algebraically:
- visually: draw a few sequences that tend to zero, and see what happens when you invert them; try to find special cases
- algebraically: knowing A_n tends to zero gives you some information about |A_n|; can you turn this into information about |1/A_n|?

 

[cloud360]

if A_n tends to zero is means that we can find the sum of A_n. it is some finite number. so 1/A_n would definitely not go to infinity. let say A_n was 25, then 1/25 is not infinity.

so the answer is "that it is not necessarily the case that if An==>0 as n==> infinity, then 1/An==>infinity as n==>infinity"

am i correct?

 

[HallsofIvy]

if A_n tends to zero is means that we can find the sum of A_n. it is some finite number. so 1/A_n would definitely not go to infinity. let say A_n was 25, then 1/25 is not infinity.

so the answer is "that it is not necessarily the case that if An==>0 as n==> infinity, then 1/An==>infinity as n==>infinity"

am i correct?

No, you are not correct. The fact that A_n goes to 0 does NOT "mean that we can find the sum of A_n". For example, A_n= 1/n goes to 0 as n goes to infinity, but \sum A_n does NOT converge (by the integral test).

A_n going to 0 is a "necessary" condition for convergence, not a sufficient condition.

 

[nonequilibrium]

You confuse me:

- "if A_n tends to zero is means that we can find the sum of A_n." Why? Take A_n = 1/n. This goes to zero, but \sum \frac{1}{n} = + \infty

- "so 1/A_n would definitely not go to infinity." even if your previous statement were true, I'm not sure how this would follow. What is the connection between \sum A_n < \infty and 1/A_n NOT going to infinity? Again, take A_n = 1/n². It can be proven that this sum will converge, but obviously 1/A_n = n² will go to infinity

- "let say A_n was 25, then 1/25 is not infinity." What do you mean with 25? The sequence A_n = 25 for each n? This doesn't go to zero.

EDIT: you seem to be VERY confused about the basics of sequences. Have you reviewed your theory before trying these exercises? I suggest you start there.

 

[cloud360]

Because i thought if a series converges it means we can find sum to infinity. guess i was wrong. will have to go back to basics
In the mean time, can you please kindly tell. does this mean the answer is "yes, if An==>0 as n==> infinity, then 1/An==>infinity as n==>infinity"

i think the answer is yes, because if A_n goes to 0, then visualizing this would mean A_n is like some fractions e.g 0.9,0.8,0.7,0.6...e.t.c

then 1/A_n would go to infinity !

(i am trying to think of this in terms of p series, guess this is wrong way)

 

[nonequilibrium]

Sticking to the basics is a good idea. Things as p-series is not necessary, it's "too advanced" for questions as these, and you should always try to find the most elegant way to understand things.

So you seem to have gotten the visual realization that if A_n goes to zero, the inverse of the series must go to infinity. Good! Any idea how you can prove this, so you can be sure? (it's possible in only one line! Tip: for every epsilon there is a n so that for every N > n: |A_n| < epsilon; this constitutes half the line I was speaking of. What does this tell you about the inverse?)