0
$\begingroup$

I want to calculate the infinite sum of the series below.

$$\sum_{k=2}^{\infty} \frac{2^{2k-1}}{5^{k+3}}$$

But unfortunately, I have no idea how to even start. Can I somehow use the formula of geometric series?

$$\sum_{k=2}^{\infty} ar^{k} = \frac{a}{1-r}$$

If I cannot, how should I solve the problem?

Thanks.

$\endgroup$
3
  • 2
    $\begingroup$ First remove a factor $\frac 1{250}=2^{-1}5^{-3}$, then see what you can make of it... $\endgroup$ – abiessu Sep 29 '18 at 14:13
  • 3
    $\begingroup$ Careful, I think the $\frac{a}{1-r}$ formula requires that the sum starts from $k=0$? $\endgroup$ – Jeppe Stig Nielsen Sep 29 '18 at 14:25
  • $\begingroup$ You are right, Jeppe. $\endgroup$ – werck Sep 29 '18 at 16:20
4
$\begingroup$

Certainly, you just got to be clever. Notice:

\begin{align*} \frac{2^{2k-1}}{5^{k+3}} &= \frac{2^{2k}}{5^k} \cdot \frac{2^{-1}}{5^3} \\ &= \frac{1}{250} \cdot \frac{4^{k}}{5^k} \end{align*}

Can you take it from here?

$\endgroup$
3
  • $\begingroup$ Oh, okay, of course. Thank you. $\endgroup$ – werck Sep 29 '18 at 14:28
  • $\begingroup$ Unfortunately, this answer is not complete, as the formula can't be directly used if the lower sum is not 0. The whole thing has to be multiplied by $(\frac{4}{5})^2$ before using the formula. $\endgroup$ – werck Sep 29 '18 at 16:21
  • $\begingroup$ True - but the OP should know to peel terms or change variables to make the formula useful. $\endgroup$ – Sean Roberson Sep 29 '18 at 16:22

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.