Question

How do you solve this?
I know you have to use the e^x identity but idk how to use it

Answer 1

Recall the exponential function's power series,

`e^x=\displaystyle\sum_(n=0)^\infty(x^n)/(n!)`

The given sum has
`n+1` instead of
`n`. We can adjust the given sum to get something usable. Notice that in the series expansion for
`e^x`, the degree of
`x` starts at 0, while in the given series, the degree starts at 1. We can actually rewrite the given series so that it starts at a different index:

`\displaystyle\sum_(n=0)^\infty(x^(n+1))/((n+1)!)=\sum_(n=1)^\infty(x^n)/(n!)`

Then this is exactly
`e^x` without the
`n=0` term. When
`n=0`, we have
`(x^0)/(0!)=1`, so

`\displaystyle\sum_(n=0)^\infty(x^(n+1))/((n+1)!)=-1+\sum_(n=0)^\infty(x^n)/(n!)=e^x-1`