Find step response for LTI system represented by the following impulse response? (i) $ h[n]=\left(\dfrac{1}{3}\right)^nu[n]+\left(-\dfrac{1}{3}\right)^nu[-n-1]$

here's how i tried to solve :

since we know, step response = running sum of impulse response 
i.e, $$s[n]=\sum_{k=-\infty}^{n} h[k] $$
we can write


  $$ s[n]=\sum_{k=-\infty}^{n}\left(\dfrac{1}{3}\right)^k\left[u[k]+(-1)^ku[-k-1]\right]$$
  $$s[n]=\sum_{k=-\infty}^{-1}\left(-\dfrac{1}{3}\right)^k +\sum_{k=0}^{n}\left(\dfrac{1}{3}\right)^k$$
  $$s[n]=(-\infty .......+3^4-3^3+3^2-3)+\dfrac{3}{2}.\left(1-\dfrac{1}{3}^{n+1}\right)$$
  $$s[n]=\left(-3+3^2-3^3+3^4......-\infty\right)+\left[\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{1}{3}\right)^n\right]$$and since sum within parethesis is diverging therefore, 
  $$s[n]=\infty$$
  but, answer in the book is given as $$s[n]=\left(2^{n+1}-1\right)u[n]$$


above s[n] denotes step response

please tell me where i'm doing wrong(or if i'm doing conceptual mistake)
,and provide solution if possible ....thank you.
The given impulse response is unbounded for $n\rightarrow -\infty$, and, consequently, the corresponding step response is also unbounded for all $n$, as you've shown. The step response given in the answer corresponds to a totally different impulse response:

$$h[n]=2^nu[n]\tag{1}$$

which is also unbounded for $n\rightarrow\infty$, but since the corresponding system is causal, the step response assumes finite values for finite $n$.