Sum of an infinite series(not quite geometric)

[GNelson]

Homework Statement

Determine the sum of the series:

\sum^{infinity}_{K=10} \frac{7}{e^(3k+2)}

Homework Equations

The Attempt at a Solution

limit n->infinity of sn=\sum^{n}_{K=10} \frac{7}{e^(3k+2)}=\frac{7}{e^(32)}+\frac{7}{e^(35)}...\frac{7}{e^(3n+2)}

This series does not exactly fit a geometric series or any partial fraction I can reduce. After this point I am stuck, any hints are greatly welcome.

My alternative method I attempted was to try to re-write it in such a way it was geometric., where f=1 is when k=10
that is that \sum^{infinity}_{K=10} \frac{7}{e^(3k+2)}= \sum^{infinity}_{f=1}\frac{7}{e^(32)}*(\frac{1}{e^(3)})^(f-1) which when simplfied I got = \frac{7}{e^(31)-1}

 

[John Creighto]

rewriting it as a geometric series makes sense but how did 3k+2 become 32?

 

[quasar987]

Have you investigated the consequences of e^{3k+2} = e^{2}(e^3)^k?

P.S. You want to write [ t e x ] \sum_{k=10}^{\infty} [ \ t e x ]

and

[ t e x ] e^{3k+2} [ t e x ]

for it to come out correctly.

 

[HallsofIvy]

quasar987's point is that this is not "not quite" but exactly a geometric series. There is a standard formula for the sum of a geometric series.

 

[GNelson]

My rational in the re-write was to write it in such a way that the terms of the sequence would be identical, in the orignal form we obtain 7/e^32+7/e^35 ect.. which was the only method of solution that I could see that had a chance of working. I tested each k value against f values at f=1 against k=10, they produce identical terms for every valid k the only difference being the new series starts at 1 rather than 10 but the way I see it we are summing an infinite amount of terms,I was curious if this was mathematically valid.

 

[Nevuse]

Correct me if I'm wrong (I'm more of a mathematician really), and I'm not all that familiar with your notation, but it is a convergent geometric progression.

It can be rearranged as sum to infinity of 7e^-2 * (e^-3)^k

which is a geometric progression with the first term (a) of 7e^-32

and increases by a factor (r) of e^-3

Using the geometric progression formula of summation of series i.e. sum=a/(1-r)

sum= (7e^-32)/(1-e^-3)

Isn't it?

 

[HallsofIvy]

Why would being "more of a mathematician" imply that you would be wrong about a math problem?

 

[Nevuse]

Why would being "more of a mathematician" imply that you would be wrong about a math problem?

In itself, it wouldn't. But, as I mentioned before I haven't used notation quite like that, so my interpretation could be wrong. (I assumed that the notation was specific to physics)

 

[dynamicsolo]

rewriting it as a geometric series makes sense but how did 3k+2 become 32?

That's intended to be the staged exponential (e^3)^{2} . One of the things TeX doesn't seem to do is double exponents. The intent of the expression is clear, if the notation may not be...

 

[Dick]

That's intended to be the staged exponential (e^3)^{2} . One of the things TeX doesn't seem to do is double exponents. The intent of the expression is clear, if the notation may not be...

No, it really is e^32. It's e^(3k+2) and the sum starts at k=10. That's e^(thirty two).

 

[dynamicsolo]

Oops, right: I misled myself in reading some of the other posts...