Infinite Series: sigma (2^n)+1/(2^n+1)

  • Thread starter anderma8
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  • #1
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i'm not quiet sure how to attack this problem:

sigma (2^n)+1/(2^(n+1))
n->1

If I start plugging in #'s for n, then I get:

n=1: 3/4
n=2: 5/8
n=3: 9/16...

by this method, I see that it's going to 1/2, but I need another way to 'see' that. Any suggestions?
 

Answers and Replies

  • #2
quasar987
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There are very few series for which we have an exact formula for the sum. One for which we do is the geometric series. Can you find a way to express the above series as, say, a sum of two geometric series?
 
  • #3
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(2^n)/(2^(n+1)) + 1/(2^(n+1))

This gives me 1/2 + 1/2^(n+1) which eventually goes to zero.... :-)

hence - my answer!
 
  • #4
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This series is exactly term by term: [tex]\frac{2^2+1}{2^(n+1)}=\frac{1}{2}+\frac{1}{2^(n+1)}[/tex]

Oh! I see anderma8 got their first, before I could figure out the latex.
 
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  • #5
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robert Ihnot - Thanks for your input though! I guess I will need to substitute numbers to get the 1/2 and see that the other side goes to 0!

Thanks! It's starting to make sense!
 
  • #6
quasar987
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If I start plugging in #'s for n, then I get:

n=1: 3/4
n=2: 5/8
n=3: 9/16...

This is just the terms in the argument of the series. You're not asked to find the limit of the argument; you're asked to find out what the series sums to. What is a series anyway? It's a sequence of partial sums. So the sequence of which you're trying to find the limit of actually has as first 3 terms

n=1: 3/4
n=2: 3/4 + 5/8
n=3: 3/4 + 5/8 + 9/16...

(2^n)/(2^(n+1)) + 1/(2^(n+1))

This gives me 1/2 + 1/2^(n+1) which eventually goes to zero.... :-)

hence - my answer!

hence nada.

What you've done is you've broken the nasty looking initial formula for the argument into 1/2 + 1/2^{n+1}. You still need to evaluate

[tex]\sum_{n=1}^{\infty}\left(\frac{1}{2}+\frac{1}{2^{n+1}}\right)[/tex]

and give a rigourous argument as to why it diverges.


P.S. I hallucinated in my first post; it's not a sum of two geometric series.
 
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  • #7
HallsofIvy
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i'm not quiet sure how to attack this problem:

sigma (2^n)+1/(2^(n+1))
n->1

If I start plugging in #'s for n, then I get:

n=1: 3/4
n=2: 5/8
n=3: 9/16...

by this method, I see that it's going to 1/2, but I need another way to 'see' that. Any suggestions?

First, what you wrote is pretty close to non-sense:
"sigma (2^n)+1/(2^(n+1))
n->1"
What in the world could "n->1" mean? I assume you mean that the sum is from 1 to infinity. Second, I feel sure you mean (2^n+ 1)/2^(n+1), not what you wrote. Finally, you do NOT get
"n=1: 3/4
n=2: 5/8
n=3: 9/16... "
Because that is a sum you "get" 3/4, 3/4+ 5/8= 11/8, 3/4+ 5/8+ 9/16= 21/16, ...

As several people have told you, the individual terms are (2^n+1)/2^(n+1)= 1/2+ 1/2^(n+1) so that the sequence of terms goes to 1/2 in the limit. Since that sequence does not go to 0, the sum itself does not converge.

(If you really did mean "sigma 2^n+ 1/(2^(n+1))", that's even worse since even the sequence of terms diverges!)
 
  • #8
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quasar987 & HallsofIvy,

It's late where I am, but I wanted to thank you both. I am sure I am confusing several words in my posts and I'm not ashamed to admit it. That's at least how I learn! Thanks for the clarification. I have been looking at this stuff for quite a bit of time this evening and will continue this tomorrow. I'll be printing out ur posts and try and internalize them. Again, thanks for your posts!
 

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