# series convergence/divergence tests (calc II)

• 04-12-2006
lschmidt
series convergence/divergence tests (calc II)
Alright, so tell me if this is bunk. I have gone through every problem in my book. I can look at the series for around 30 seconds or less, and decide whether it converges or diverges.

Yet we have to use all these stupid tests...

What gives?
• 04-12-2006
jverkoey
What do you mean? Knowing integral convergence test, nth term test, alternating series test, and etc...? All these theorems are there mainly to take in to account most of the general equations you'll run in to, so they're good to know.

Also, they're the formal way of proving that the series converges/diverges. Just "knowing" that a series conv/div isn't good enough in academia.
• 04-12-2006
Sang-drax
Quote:

Originally Posted by lschmidt
Alright, so tell me if this is bunk. I have gone through every problem in my book. I can look at the series for around 30 seconds or less, and decide whether it converges or diverges.

You've misunderstood the purpose of the problems in your book. Their purpose is to help you understand the theory, not to learn by heart to avoid the theory.
In real life/other courses you won't have any benefits from being able to look at a series and tell if it converges. If you learn all the 'stupid tests' though, you'll gain useful knowledge of calculus.
• 04-12-2006
Rashakil Fol
Does this series converge? You have 30 seconds.

|cos(1)|/1 + |cos(2)|/2 + |cos(3)|/3 + ...

1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 + ...
• 04-13-2006
lschmidt
I hope you are kidding.
• 04-13-2006
jverkoey
No, we never kid here.
• 04-13-2006
Rashakil Fol
Okay, so both of those fairly obviously diverge. You probably couldn't prove it easily, though, unless you had the right tools available. And if somebody sat there convinced that the sum of the reciprocals of primes converged, how would you convince them?

In math, you can't get away with going by your gut feeling, because at one point in time or another, you'll be wrong. For example, the sum of

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + - ...

is log(2). Well all I'm going to do is move a few terms around. What's the value of the sum,

1 - 1/2 + 1/3 + 1/5 - 1/4 + 1/7 + 1/9 - 1/6 + 1/11 + 1/13 - 1/8 + + - ...

And is it possible to take a solid ball, split it into ten subsets of points, and then only by moving those subsets around rigidly, form a solid ball with half the radius? The gut feeling says no.
• 04-15-2006
jafet

SUM (1/x)

Where x are the numbers without the digit 9?

Or
1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + 21/256...
?

Besides, math isn't about gut feelings. It's about formal, solid, axiomatic proof.
• 04-15-2006
Rashakil Fol
Interesting ones!
Quote:

Originally Posted by jafet
SUM (1/x)

Where x are the numbers without the digit 9?

Is less than 10*(1+1/2+1/3+1/4+1/5+1/6+1/7+1/8).

Quote:

Originally Posted by jafet
Or
1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + 21/256...
?

Is less than phi^2.
• 04-15-2006
Sang-drax
Quote:

Originally Posted by Rashakil Fol
Is less than 10*(1+1/2+1/3+1/4+1/5+1/6+1/7+1/8).

Why?
• 04-15-2006
Rashakil Fol
You can put parentheses around pieces of the series, grouping together the fractions whose denominators have the same number of digits. E.g.

(1/1 + ... + 1/8) + (1/10 + ... + 1/88) + (1/100 + ... + 1/888) + ...

Since 1/(10n + r) < 1/(10n), for 1 <= r <= 8, we know that

((1/10 + 1/11 + 1/12 + ... + 1/18) + (1/20 + 1/21 + 1/22 + ... + 1/28) + ... + (1/80 + 1/81 + 1/82 + ... + 1/88)) <
((1/10 + 1/10 + 1/10 + ... + 1/10) + (1/20 + 1/20 + 1/20 + ... + 1/20) + ... + (1/80 + 1/80 + 1/80 + ... + 1/80)) =
9*(1/10 + 1/20 + ... + 1/80) =
(9/10) * (1/1 + 1/2 + ... + 1/8).

This argument applies for the rest of the parenthesized groups. So the sum for each parenthesized group is less than 9/10 the sum of the previous parenthesized group. From there we know that the overall sum is less than that of a geometric series with ratio 9/10, whose first term is 1+1/2+...+1/8.
• 04-15-2006
gcn_zelda
How does knowing whether a series converges or diverges come into play in "the real world"?

I'm not doubting that it does. I just can't think of any occasion in which it would be of any assistance.
• 04-15-2006
Rashakil Fol
If you're trying to invent an algorithm that approximates something, you'll want to know whether it approximates that thing well, and how quickly it can come up with that approximation. Your approximation method might only work for certain inputs, so you'd need to know which inputs these work on. Or it might only work quickly for certain inputs. With certain restrictions of input, you can have a guarantee of how long it takes to get a certain degree of approximation.

The 'something' could be a real number, a picture, a music file, ....