Compute $f(x) = \sum_{k = 1}^{\infty} \Bigg(\frac{1}{(k-1)!} + k\Bigg)x^{k-1}$

Question

Compute $f(x) = \sum_{k = 1}^{\infty} \Bigg(\frac{1}{(k-1)!} + k\Bigg)x^{k-1}$

Approach

I'm not exactly sure how to do this, but just shooting around ideas due to this question appearing in a chapter on power series and uniform convergence, by idea would be to possibly find the radius of convergence first. From there perhaps I could reduce the set I could work on and I might have a series that I know converges to use as a bound and approximation for this series. Other than that nothing else comes to mind at the moment.

Answer 1

This sum is $$ \sum_{k = 0}^{\infty} \Bigg(\frac{1}{k!} + k + 1\Bigg)x^{k} = \sum_{k = 0}^{\infty} \frac{x^k}{k!} + \sum_{k = 0}^{\infty}kx^k + \sum_{k = 0}^{\infty}x^k. $$ First sum is just $e^x$. Third sum is just geometric progression. In order to compute the second sum take the derivative of the function $$ \frac{1}{1-x} = \sum_{k=0}^\infty x^k. $$

Answer 2

$$\int_0^x f(t)\mathrm{d}t=\sum_{k=1}^\infty\left(\frac1{k!}+1\right)x^k=\sum_{k=1}^\infty\left(\frac{x^k}{k!}+x^k\right)=e^x-1+\frac{x}{1-x}$$ Hence we have that $$f(x)=\frac{\mathrm{d}}{\mathrm{d}x}\int_0^x f(t)\mathrm{d}t=e^x+\frac1{(1-x)^2}$$ also the radius of convergence is $1$ which follows from the radius of convergence of the power series $\sum x^k$.