(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.