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.

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.