$$\sum_{n=1}^\infty \frac{5(-2)^n-3^{(n+1)}}{5^n}$$

tried using geometric series, but unable to factor out the constant due to the minus sign.

Also tried using ratio test but I got stuck halfway.

  • $\begingroup$ Did you happen to mean $$(-2)^n\text{ or }-(2^n)$$? $\endgroup$ – Simply Beautiful Art Oct 11 '17 at 1:41
  • $\begingroup$ (−2)^n sorry for bad formatting $\endgroup$ – J. Wilshere Oct 11 '17 at 1:46
  • 2
    $\begingroup$ Hint:$$\sum_{n=1}^\infty\frac{a\cdot b^n-c^{n+1}}{d^n}=\sum_{n=1}^\infty\left[a\cdot\left(\frac bd\right)^n-c\cdot\left(\frac cd\right)^n\right]$$And you know that:$$\sum_{n=1}^\infty ar^n=\frac{ar}{1-r},~|r|<1$$ $\endgroup$ – Simply Beautiful Art Oct 11 '17 at 1:48
  • $\begingroup$ is it okay to split the series into two, and calculate each limit separately, followed by subtracting the second term by the first term. if so, i think i got it $\endgroup$ – J. Wilshere Oct 11 '17 at 1:59
  • $\begingroup$ Yes, this follows from the fact that you can split a series if each part is absolutely convergent. (Or you could just check the partial sums, which is the elementary justification) $\endgroup$ – Simply Beautiful Art Oct 11 '17 at 2:03

It's $$5\sum_{n=1}^\infty\left(-{2\over5}\right)^n-3\sum_{n=1}^\infty\left({3\over5}\right)^n=5{-{2\over5}\over1-\bigl(-{2\over5}\bigr)}-3{{3\over5}\over1-{3\over5}}=\ldots$$


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