Question

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!}

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asked 2021-03-02
Differentiate the power series for \(f(x)=xe^x\). Use the result to find the sum of the infinite series
\(\sum_{n=0}^\infty\frac{n+1}{n!}\)

Answers (1)

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.
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