Sum of an infinite series(not quite geometric)

I was thinking that the expression was 7/((e^3)^k+2), not (7/e^3)^k+2.And so I think I now agree with quasar.In summary, the series given by \sum^{infinity}_{K=10} \frac{7}{e^(3k+2)} can be rewritten as a geometric series with the first term (a) of 7e^-32 and a common ratio (r) of e^-3. Using the formula for the sum of a geometric series, the sum can be calculated as (7e^-32)/(1-e^-3).
  • #1
GNelson
9
0

Homework Statement



Determine the sum of the series:

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

Homework Equations


The Attempt at a Solution

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

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 [tex]\sum[/tex][tex]^{infinity}_{K=10}[/tex] [tex]\frac{7}{e^(3k+2)}[/tex]= [tex]\sum[/tex][tex]^{infinity}_{f=1}[/tex][tex]\frac{7}{e^(32)}[/tex]*([tex]\frac{1}{e^(3)}[/tex])^(f-1) which when simplfied I got = [tex]\frac{7}{e^(31)-1}[/tex]
 
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  • #2
rewriting it as a geometric series makes sense but how did 3k+2 become 32?
 
  • #3
Have you investigated the consequences of [tex]e^{3k+2} = e^{2}(e^3)^k[/tex]?

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.
 
  • #4
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.
 
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  • #5
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.
 
  • #6
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?
 
  • #7
Why would being "more of a mathematician" imply that you would be wrong about a math problem?:biggrin:
 
  • #8
HallsofIvy said:
Why would being "more of a mathematician" imply that you would be wrong about a math problem?:biggrin:

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)
 
  • #9
John Creighto said:
rewriting it as a geometric series makes sense but how did 3k+2 become 32?

That's intended to be the staged exponential [tex](e^3)^{2}[/tex] . 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...
 
  • #10
dynamicsolo said:
That's intended to be the staged exponential [tex](e^3)^{2}[/tex] . 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).
 
  • #11
Oops, right: I misled myself in reading some of the other posts...
 

Related to Sum of an infinite series(not quite geometric)

1. What is an infinite series?

An infinite series is a sum of an infinite number of terms. Each term is added to the previous one, and the sum can either converge to a finite value or diverge to infinity.

2. How is the sum of an infinite series calculated?

The sum of an infinite series can be calculated using various methods, such as rearranging the terms to form a geometric series, using the ratio test or the integral test, or using the properties of convergent series.

3. What is the difference between a geometric series and a non-geometric series?

A geometric series has a constant ratio between consecutive terms, while a non-geometric series does not have a constant ratio. Geometric series are easier to work with as their sum can be calculated using a simple formula, while the sum of a non-geometric series can be more complex to determine.

4. Can an infinite series have a sum?

Yes, an infinite series can have a sum as long as it converges. If the series diverges, it does not have a finite sum.

5. What is the significance of the sum of an infinite series in mathematics?

The sum of an infinite series is a fundamental concept in mathematics and has many applications in calculus, physics, and engineering. It allows us to model real-world situations and make precise calculations, and it also helps in understanding the behavior of functions and sequences.

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