- #1

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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?

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- Thread starter anderma8
- Start date

- #1

- 35

- 0

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?

- #2

quasar987

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- #3

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

hence - my answer!

- #4

- 1,056

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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.

Oh! I see anderma8 got their first, before I could figure out the latex.

Last edited:

- #5

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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...

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.

Last edited:

- #7

HallsofIvy

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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

(If you really did mean "sigma 2^n+ 1/(2^(n+1))", that's even worse since even the

- #8

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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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