0
$\begingroup$

Determine the sum of the following geometric series.

I got confused here because I'm used to starting with $k=1$ and now suddenly it's $k=5$. Is there any difference?

$$\sum_{k=5}^ \infty \left(\frac{e}{\pi}\right)^{k-1} $$

$\endgroup$
  • $\begingroup$ Just assume that the sum starts at $1$, determine the solution and then subtract the sum from $1$ to $4$ from the result. Du you understand what I mean? $\endgroup$ – Moritz Sep 12 '16 at 13:49
  • $\begingroup$ determine the solution, yes got it, subtract sum from 1 to 4, didn't get it, could you be more specific $\endgroup$ – Teeban Sep 12 '16 at 13:51
  • $\begingroup$ $\sum_{k=1}^\infty a_k = \sum_{k=1}^4 a_k + \sum_{k=5}^\infty a_k$ $\endgroup$ – Nicholas Stull Sep 12 '16 at 13:53
  • $\begingroup$ @Teeban: Look at the hint from Nicholas Stull. $\endgroup$ – Moritz Sep 12 '16 at 13:54
0
$\begingroup$

you can use the geometric series $$\sum_{k=0}^{\infty }x^k=\frac{1}{1-x} \quad{|x|<1}$$ so $$\sum_{k=5}^ \infty (\frac{e}{\pi})^{k-1}=\sum_{k=0}^ \infty (\frac{e}{\pi})^{k+4}=(\frac{e}{\pi})^4\sum_{k=0}^ \infty (\frac{e}{\pi})^{k}=(\frac{e}{\pi})^4\frac{1}{1-\frac{e}{\pi}}$$

$\endgroup$
  • $\begingroup$ How , Can you explain please $\endgroup$ – Teeban Sep 12 '16 at 13:44
  • 1
    $\begingroup$ Maybe get rid of the phony = signs? $\endgroup$ – Did Sep 12 '16 at 13:44
  • $\begingroup$ $$\sum_{k=1}^ \infty ar^{k-1}$$ is the formula for sum, if k=0 how can i find the sum $\endgroup$ – Teeban Sep 12 '16 at 13:48
  • 1
    $\begingroup$ can you correct your post please? $\endgroup$ – Dr. Sonnhard Graubner Sep 12 '16 at 13:48
  • $\begingroup$ done the correction $\endgroup$ – Teeban Sep 12 '16 at 13:50
0
$\begingroup$

Hint: $$\sum_{k=a}^\infty f(k) = \sum_{k=1}^\infty f(k) - \sum_{k=1}^{a-1}f(k)$$ or more generally if $a < b < c$ $$\sum_{k=b}^c f(k) = \sum_{k=a}^c f(k) - \sum_{k=a}^{b-1}f(k)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.