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?


  • 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

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?

  • $\begingroup$ 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

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