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

[anderma8]

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?

 

[quasar987]

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?

 

[anderma8]

(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!

 

[robert Ihnot]

This series is exactly term by term: $$\frac{2^2+1}{2^(n+1)}=\frac{1}{2}+\frac{1}{2^(n+1)}$$

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

 

[anderma8]

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!

 

[quasar987]

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

$$\sum_{n=1}^{\infty}\left(\frac{1}{2}+\frac{1}{2^{n+1}}\right)$$

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.

 

[HallsofIvy]

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!)

 

[anderma8]

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!