Taylor development series

Question

I don't quite understand how to get from:
$$ 3e^{t/3} = \sum_{n=0}^{\infty} \frac{1}{3^{n-1}n!}t^n$$

to:

$$ 3e^{t/3} -3 - t = \sum_{n=2}^{\infty} \frac{1}{3^{n-1}n!}t^n $$

That is what my teacher did, but I can't figure out where did the -3 and -t go.

Answer 1

The expression$$\sum_{n=0}^{\infty}\frac1{3^{n-1}n!}t^n$$means$$\frac1{3^{-1}}+t+\frac1{3\times2}t^2+\frac1{3^2\times3!}t^3+\cdots$$Its first two terms are $3+t$ and therefore, since$$3e^{t/3}=\sum_{n=0}^{\infty} \frac{1}{3^{n-1}n!}t^n=\frac1{3^{-1}}+t+\frac1{3\times2}t^2+\frac1{3^2\times3!}t^3+\cdots,$$you have$$3e^{t/3}-3-t=\frac1{3\times2}t^2+\frac1{3^2\times3!}t^3+\cdots=\sum_{n=2}^\infty\frac1{3^{n-1}n!}t^n.$$

Answer 2

Simply, $$\sum_{n=0}^\infty a_n=a_0+\sum_{n=1}^\infty a_n=a_0+a_1+\sum_{n=2}^\infty a_n$$ and

$$\sum_{n=0}^\infty a_n-a_0-a_1=\sum_{n=2}^\infty a_n.$$