- #1

- 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]

Last edited: