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

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

quasar987
Science Advisor
Homework Helper
Gold Member
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?

(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!

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.

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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!

quasar987
Science Advisor
Homework Helper
Gold Member
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.

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HallsofIvy
Science Advisor
Homework Helper
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!)

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!