This is the series:
$$\sum_{k=1}^{\infty}ke^k$$
I need to calculate Max and Min with integral comparison.
$$\int_1^nxe^x dx \leq \sum_{k=1}^n ke^k \leq \int_1^{n+1}xe^xdx$$ Calculate the indefinite integral:
$$ \int xe^xdx = e^x + xe^x = e^x(x-1) + C$$ $$ \big[ e^x(x-1) \big]_1^n \leq \sum_{k=1}^n ke^k \leq \big[ e^x(x-1) \big]_1^{n+1}$$
Calculate the Min:
$$\big[ e^x(x-1) \big]_1^n = e^n(n-1) - e^1(1-1) = e^n(n-1)$$
Calculate the Max:
$$\big[ e^x(x-1) \big]_1^{n+1} = e^{n+1}(n+1-1) - e^1(1-1) = e^{n+1}(n+1-1) = ne^{n+1}$$
So:
$$ e^n(n-1) \leq \sum_{k=1}^{\infty}ke^k \leq ne^{n+1}$$
Now, this is a geometric series with $|e^k|>1$ so it should diverge.
If I calculate the limit:
$$ \lim_{x\to\infty} ke^k = +\infty$$
So the series diverges, correct?