# Differentiate the power series for f(x)=xe^x. Use the result to find the sum of the infinite series sum_{n=0}^inftyfrac{n+1}{n!}

Question
Series
Differentiate the power series for $$f(x)=xe^x$$. Use the result to find the sum of the infinite series
$$\sum_{n=0}^\infty\frac{n+1}{n!}$$

2021-03-03
Consider the function $$f(x)=xe^x$$
Differentiate the above function to get,
$$f'(x)=\frac{d}{dx}(xe^x)$$
$$=xe^x+e^x$$
$$=\sum_{n=0}^\infty\frac{(n+1)x^n}{n!}$$
Substitute 1 for x in the above series,
$$1e^1+e^1=\sum_{n=0}^\infty\frac{(n+1)(1)^n}{n!}$$
$$2e=\sum_{n=0}^\infty\frac{(n+1)}{n!}$$
$$\sum_{n=0}^\infty\frac{(n+1)}{n!}\approx5.4363$$
Thus, the sum of the infinite series is approximately 5.436.

### Relevant Questions

a) Find the Maclaurin series for the function
$$f(x)=\frac11+x$$
b) Use differentiation of power series and the result of part a) to find the Maclaurin series for the function
$$g(x)=\frac{1}{(x+1)^2}$$
c) Use differentiation of power series and the result of part b) to find the Maclaurin series for the function
$$h(x)=\frac{1}{(x+1)^3}$$
d) Find the sum of the series
$$\sum_{n=3}^\infty \frac{n(n-1)}{2n}$$
This is a Taylor series problem, I understand parts a - c but I do not understand how to do part d where the answer is $$\frac72$$
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