# Infinite Series

1. Nov 21, 2007

### danni7070

[Solved] Infinite Series

1. The problem statement, all variables and given/known data
Find the sum of the given series, or show that the series diverges.

_∞
∑_(k=0) (2^(k+3))/(℮^(k-3))

I hope this is not confusing, and it would be great if someone knows about some site where you can write equations easily online.

2. Relevant equations

I don't know any which can help me here.

3. The attempt at a solution

I just see what happens when k = 0, 1, 2, 3..... inf and see what happens. But to find the sum I'm totally lost. Could somebody get me started here?

Last edited: Nov 22, 2007
2. Nov 21, 2007

### Phred101.2

k can't be < 3, for a start...

3. Nov 21, 2007

### danni7070

Why not?

4. Nov 21, 2007

### HallsofIvy

Staff Emeritus
That's actually just a straightforward geometric series! $2^{k+3}= 8 2^k$ and $e^{k-3}= e^{-3} e^k$ so your series is just
$$\Sum_{k=0}^\infty (8e^{-3})(2e)^k[/itex] That's a geometric series with $a= 8e^{-3}$ and $r= (2e)^k$. 5. Nov 21, 2007 ### danni7070 Isn't it suppose to be [tex] 8e^3$$ ?

$$\frac{2^3*2^k}{e^-3*e^k}$$

$$\frac{8}{e^-3}*\frac{2^k}{e^k}$$

$$8*\frac{1}{e^-3} * \frac{2^k}{e^k}$$

$$8e^3 * (\frac{2}{e})^k$$

Last edited: Nov 21, 2007
6. Nov 21, 2007

### Kreizhn

Yes, you are indeed correct. I believe HallsofIvy may have just mistaken the fact that we were dividing by $$e^{k-3}$$ by multiplying. As a matter of fact, I made the same mistake when reviewing your question. It's easy to see why it wouldn't make sense otherwise, since the sum would then diverge to infinity.

Other than the sign error though, HallsofIvy is exactly correct. Since $$\frac{2}{e} < 1$$ you can now just treat it as the usual infinite geometric series.

7. Nov 21, 2007

### Phred101.2

After much calculation and substitution (I used my HP49G+), I get:

$$(1- \frac { \sum_{k=3}^\infty \frac {2^{(k+3)}} { e^{k-3}} } {|\sum_{k=3}^\infty \frac {2^{(k+3)}} { e^{k-3}} | } )\frac \pi 2$$

If $$\sum_{k=3}^\infty \frac {2^{(k+3)}} { e^{k-3}}$$ is substituted with, say 'u', then it's:

$$(1- \frac {u} {|u |} )\frac \pi 2$$

This fine example of algebraic manipulation, does not allow k to be < 3, maybe because this would have an undefined denominator in those fractions...

However, solving the expression: $$\frac {2^{(k+3)}} { e^{k-3}}$$ gives:$$\frac {8e^3 e^k ln 2 } {e^k }$$

So, substituting this back into the original: $$\sum_{k=0}^\infty \frac {8e^3 e^k ln 2 } {e^k }$$

P.S. Sorry if my claim that k must be >= 3 has confused people; my calc won't let me assign 0,1, or 2 to the first eqn. if it is set to arithmetic (rather than algebraic mode), if that explains it any further...

Last edited: Nov 21, 2007
8. Nov 22, 2007

### danni7070

Ok, this one is solved. Thanks everyone for explaining.

$$\sum_{k=0}^\infty \frac {2^{(k+3)}} { e^{k-3}}$$

$$8e^3\sum_{k=3}^\infty (\frac{2}{e})^k$$

$$\frac {8e^3}{1-\frac{2}{e}}$$

$$\frac {8e^4}{e-2}$$

And the answer from the book agrees! This feels great.