A series problem

[bioshock]

what is the sum of this infinite series ?

$\displaystyle 1+1/(3*2^2)+1/(5*2^4)+1/(7*2^6)+................$

i see the General term(actually from 2nd term) as $\displaystyle 1/((2n+1)*2^{2n})$

i need a process to get sum. there are many problems similar to this and i need to solve them.
and oh this is my first post so if i have done something wrong please forgive me.
thanks

 

[mmm4444bot]

… i see the common term … as $\displaystyle 1/((2n+1)*2^2n)$ …

I would call that the general term, and LaTex requires curly braces around the exponent 2n, for proper formatting. Click the

button, to see LaTex coding for the following.

$\displaystyle \frac{1}{(2n + 1) \cdot 2^{2n}} \; , \; n \ge 0$

In the absense of more self-review (on my part) than I have time for right now, the only answer to your question that I can currently give is a decimal approximation (from a numerical approach): 1.098612289.

This looks like the natural logarithm of 3, to me!

Hang tough. I'm confident that somebody else will help you with this exercise.

BTW, this post probably doesn't belong on the Pre-Algebra board, but don't be concerned because its placement here is really not a big deal. If the regulars around here do as I do, then we use the 'View active topics' link to peruse new posts, so your post will be seen, regardless. We also have moderators, for moving threads around the boards (if they see fit).

 

[bioshock]

OK fixing the post.Thanks for the quick reply. I will be waiting for the answer

 

[Denis]

i see the General term(actually from 2nd term) as $\displaystyle 1/((2n+1)*2^{2n})$

Less confusing if "from the 1st term".
1 / [x * 2^(x - 1)] where x = 2n - 1

 

[mmm4444bot]

Gosh, maybe people are on vacation.

PKA? Dr. Mike? Galactus? Big Glenn? Subhotosh? Is anybody there? Anyone at all? :wink:

(Maybe this exercise requires a lot of steps or explanation; that is, more steps or explanation than people want to type.)

I spent about 45 minutes reviewing (on the Internet), but I didn't get very far. Manipulating summation formulas with n as both a factor and an exponent is proving difficult, for me.

I'll e-mail a couple people I know outside of this forum, to see what they think.

 

[Denis]

Just looking at adding the 1st 4 terms...to "see" what d'heck's going on!
Re-arranging like 1 / (a * 2^n) = 2^(-n) / a, then multiplying the series by 2^2 gives:

S      =          2^0/1  +  2^(-2)/3  +  2^(-4)/5  + 2^(-6)/7
S(2^2) = 2^2/1 +  2^0/3  +  2^(-2)/5  +  2^(-4)/7

S(2^2) - S = 2^2 + [ stuck here! ] - 2^(-6)/7

S(2^2 - 1) = 2^2 + [ stuck here! ] - 2^(-6)/7

S = {4 + [ stuck here! ] - 2^(-6)/7} / 3

If I could handle that "stuck here" portion, then I'd have it made....but I can't see how

 

[Denis]

Well, good luck Bioshock.
My [stuck here] portion apparently is no gimme:
www.maa.org/editorial/euler/How%20Euler ... primes.pdf

 

[bioshock]

Thanks to everyone. lets see what i can do with it. still no solution

 

[mmm4444bot]

Nobody outside of this site responded to my e-mail inquiries. (A common lament, for me.)

They probably have too much to do already, with the start of a new school year.

Please post a method, after you get an answer. I'd like to see it.