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

How do you solve this? I know you have to use the e^x identity but idk how to use-example-1
User Sudhee
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1 Answer

3 votes

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

User Ckunder
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