Solving Infinite Series: (2^(k+3))/(℮^(k-3))

[danni7070]

[Solved] Infinite Series

Homework Statement

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.

Homework Equations

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

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?

 

[Phred101.2]

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

 

[danni7070]

Why not?

 

[HallsofIvy]

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]<br /> That&#039;s a geometric series with a= 8e^{-3} and r= (2e)^k.

 

[danni7070]

Isn't it suppose to be 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

 

[Kreizhn]

Isn't it suppose to be 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

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} &lt; 1 you can now just treat it as the usual infinite geometric series.

 

[Phred101.2]

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

(1- \frac {<br /> \sum_{k=3}^\infty \frac {2^{(k+3)}} { e^{k-3}}<br /> } <br /> {|\sum_{k=3}^\infty \frac {2^{(k+3)}} { e^{k-3}} |<br /> }<br /> )\frac \pi 2If \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 }<br />

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...

 

[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.