No.
We know that
∑n=0∞zn=11−z\sum_{n=0}^{\infin} {z^n} =\frac 1 {1-z}∑n=0∞zn=1−z1
Find derivative of it:
∑n=1∞n∗zn−1=1(1−z)2\sum_{n=1}^{\infin} {n*z^{n-1}} =\frac 1 {(1-z)^2}∑n=1∞n∗zn−1=(1−z)21
Change n to n+1
∑n=0∞(n+1)∗zn=1(1−z)2\sum_{n=0}^{\infin} {(n+1)*z^{n}} =\frac 1 {(1-z)^2}∑n=0∞(n+1)∗zn=(1−z)21
Hence, the generating function of the sequence {1,2,3...n..} is 1(1−z)2\frac 1 {(1-z)^2}(1−z)21
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