Evaluate $\sum_{n=1}^\infty(-1)^{n-1}\frac{9}{4^n}.$

Question

Evaluate $$\sum_{n=1}^\infty(-1)^{n-1}\frac{9}{4^n}.$$

My approach: using $\frac{a}{1-r}$, we have

$\frac{9}{1-\frac{1}{4}} = 12$

But the correct answer is $\frac{9}{5}$

What am I missing in solving this problem?

Answer 1

If the $r$th term is $$T_m=(-1)^{m-1}\dfrac9{4^m},$$

$\dfrac{T_{m+1}}{T_m}=-\dfrac14$

So, your $r\ne\dfrac14$

Answer 2

To relate the given expression $$\sum_{n=1}^\infty (-1)^{n-1}\frac9{4^n}$$ to the known power series $$\frac1{1-x}=\sum_{n=0}^\infty x^n, |x|<1,$$ we can write \begin{align} \sum_{n=1}^\infty (-1)^{n-1}\frac9{4^n}&=9\sum_{n=1}^\infty(-1)\left(-\frac14\right)^n\\&=-9\sum_{n=0}^\infty\left(-\frac14\right)^{n+1}\\&=-9\sum_{n=0}^\infty\left(-\frac14\right)^{n}\cdot\left(-\frac14\right)\\&=\frac94\sum_{n=0}^\infty\left(-\frac14\right)^n. \end{align}

Since $\left|-\dfrac14\right|<1,$ therefore, we can write $$\frac94\sum_{n=0}^\infty\left(-\frac14\right)^n=\frac94\left(\frac{1}{1-(-\frac14)}\right)=\frac94\cdot\frac45=\frac95$$ as desired.