# Sum of an infinite series(not quite geometric)

## Homework Statement

Determine the sum of the series:

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

## 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 = $$\frac{7}{e^(31)-1}$$

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## Answers and Replies

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

quasar987
Homework Helper
Gold Member
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
Homework Helper
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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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.

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

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
Homework Helper
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...