Series

bobbym · 2011-09-15 09:17:22

You might try using the Binet formula hoping that there would be some cancellation. Or you can try expanding the first few terms and trying to recognize the generating function.

anonimnystefy · 2011-09-15 09:22:07

i don't know any of that stuff.what about 19?

bobbym · 2011-09-15 09:27:06

19, possibly snake oil method. Or again using a table to recognize the generating function.

Also you really should know the Binet formula.

bobbym · 2011-09-16 00:06:30

Hi gAr;

Hopefully, you have a subscribe and solutions on this thread might draw you back.

anonimnystefy · 2011-09-16 04:42:35

where can i find it?

bobbym · 2011-09-16 06:26:00

It? The fountain of youth? The Oracle at Delphi? The room beneath the Sphinx? The ark of the Covenant? King Solomons gold? Brad Pitts ambition? Darth Vader's and the Cowardly Lion's courage? The opening to the alien base beneath Vegas?

gAr · 2011-09-16 18:33:02

Hi bobbym,

bobbym · 2011-09-16 21:29:53

Hi gAr;

gAr · 2011-09-16 22:08:51

bobbym · 2011-09-17 02:22:04

Hi gAr;

The best of luck to you.

gAr · 2011-09-17 15:11:38

Hi bobbym,

Thank you!

reconsideryouranswer · 2011-09-18 03:11:11

Is this convergent or divergent?

Please look at this amended pronblem. (All variables are the same now.)

Last edited by reconsideryouranswer (2011-09-19 02:47:14)

bobbym · 2011-09-18 21:03:05

Hi;

reconsideryouranswer wrote:

Since;

Has no solutions, the summand is a constant and an infinite sum of constants does not converge.

reconsideryouranswer · 2011-09-19 02:14:07

bobbym wrote:

Hi;
Since;
1)
2) Has no solutions,
the summand is a constant
and an infinite sum of constants does not converge.

1)
You meant the *limit of that expression as x --> oo*, correct?
Can you justify this?

2)

So, I don't see your first part of your message as being sufficient
to justify that the sum is divergent (not convergent).

bobbym · 2011-09-19 02:17:21

Your 2) is incorrect. That only applies when the x is part of the indices:

Which is not the same as:

The top one is adding up a sequence that is getting smaller and it converges. The second one is adding up a sequence that never changes, it diverges.

You do not need limits here. There is no value that can make the summand = 0. Therefore for all values of x the summand is a constant. It never changes, Add up an infinite bunch of constants and you have a divergent series.

reconsideryouranswer · 2011-09-19 02:48:27

reconsideryouranswer wrote:

Is this convergent or divergent?

Please look at this amended problem. (All variables are the same now.)

-----------------------------------------------------------------------
-----------------------------------------------------------------------

bobbym wrote:

"Since;
Has no solutions, ..."

There was a misunderstanding here by me,
because you didn't put this all on one line**,
and you had "Has" in the middle of a
sentence instead of "has." It appears to
be two sentences, despite "Since;[,]"
being at the beginning of the sentence.

** It is just about the visual aspect of it not being on
one line, not the correctness of it.

*** Edit:

My amended question is still open.

Last edited by reconsideryouranswer (2011-09-19 03:07:11)

bobbym · 2011-09-22 17:30:33

Hi reconsideryouranswer;

First observe:

Using the comparison test:

The RHS is a variant of the harmonic series and is known to diverge. In order for that inequality to be true then:

must be true.

Raise both sides to the n^n th power:

This is obvious for all n>=1. This now implies:

Since your sum is greater than a known series that diverges by the comparison test it too diverges.

gAr · 2011-09-26 01:06:01

22)

Show that

anonimnystefy · 2012-02-17 05:52:56

any easier sums?

gAr · 2012-02-17 05:58:58

Nope, only after all the sums here are solved!

anonimnystefy · 2012-02-17 06:01:55

but i don't know how to solve these sums.that's why i asked for some easier ones.

anonimnystefy · 2012-02-20 05:33:26

hi gAr

just wanna ask if this is ok:

Last edited by anonimnystefy (2012-02-20 05:39:18)

bobbym · 2012-02-20 10:30:53

Hi anonimnystefy;

Your #17 is not correct.

gAr · 2012-02-21 02:19:06

Hi anonimnystefy,

I did the problem bobbym's way.

anonimnystefy · 2012-02-21 03:52:07

hi guys

i looked for formulas for

but the only one i found was the one i used.

Math Is Fun Forum

#176 2011-09-15 09:17:22

Re: Series

#177 2011-09-15 09:22:07

Re: Series

#178 2011-09-15 09:27:06

Re: Series

#179 2011-09-16 00:06:30

Re: Series

#180 2011-09-16 04:42:35

Re: Series

#181 2011-09-16 06:26:00

Re: Series

#182 2011-09-16 18:33:02

Re: Series

#183 2011-09-16 21:29:53

Re: Series

#184 2011-09-16 22:08:51

Re: Series

#185 2011-09-17 02:22:04

Re: Series

#186 2011-09-17 15:11:38

Re: Series

#187 2011-09-18 03:11:11

Re: Series

#188 2011-09-18 21:03:05

Re: Series

#189 2011-09-19 02:14:07

Re: Series

#190 2011-09-19 02:17:21

Re: Series

#191 2011-09-19 02:48:27

Re: Series

#192 2011-09-22 17:30:33

Re: Series

#193 2011-09-26 01:06:01

Re: Series

#194 2012-02-17 05:52:56

Re: Series

#195 2012-02-17 05:58:58

Re: Series

#196 2012-02-17 06:01:55

Re: Series

#197 2012-02-20 05:33:26

Re: Series

#198 2012-02-20 10:30:53

Re: Series

#199 2012-02-21 02:19:06

Re: Series

#200 2012-02-21 03:52:07

Re: Series

Board footer