I'm solving a problem but I haven't been able to move forward because it gives me discordance.
Let $f$ defined by $$f(t)=n \quad t\in [n,n+1)$$ Show $$F(s)=\frac{1}{s}\sum_{n=1}^\infty e^{-sn}$$
but when I try I get to this expression $$F(S) = \frac{1-e^{-s}}{s}\sum_{n=1}^{\infty} ne^{-sn}$$ What will I be doing wrong? Thanks
Your function f can be written as $$f(t) = \sum_{n=1}^{\infty} \Theta(t-n)$$ a sum of shifted step functions. This yields the desired transformation $$F(s) = \frac{1}{s}\cdot \sum_{n=1}^{\infty} e^{-sn}$$ Your solution is also not wrong. Notice that $$ \frac{1-e^{-s}}{s}\cdot \sum_{n=1}^{\infty} ne^{-sn} = \frac{1}{s}\cdot (\sum_{n=1}^{\infty}ne^{-sn} - \sum_{n=1}^{\infty} ne^{-s(n+1)}) $$ The last two sums can be simplifed even further by adding them componentwise $$\sum_{n=1}^{\infty}ne^{-sn} - \sum_{n=1}^{\infty} ne^{-s(n+1)} = \sum_{n=1}^{\infty} (n - (n-1))\cdot e^{-sn}$$ (substitute k = n+1 and notice that k = 1 for n = 0 in the second sum on the left side) Therefore we arrive at the same solution $$F(s) = \frac{1}{s}\cdot \sum_{n=1}^{\infty} e^{-sn} = \frac{1-e^{-s}}{s}\cdot \sum_{n=1}^{\infty} ne^{-sn} $$