Find the limit as x approaches negative infinity of the sum of x^n/(n * n!) for n from 1 to infinity.

Step-by-step solution

1. 1

   Name the sum and note it converges everywhere. Put

   $S(x)=\sum_{n=1}^{\infty}\frac{x^{n}}{n\,n!}$

   The $n!$ in the denominator makes the radius of convergence infinite, so $S$ is defined for every real $x$ and may be differentiated term by term.

2. 2

   Differentiate to remove the awkward factor $n$.

   $S'(x)=\sum_{n=1}^{\infty}\frac{n\,x^{n-1}}{n\,n!}=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n!}=\frac{1}{x}\sum_{n=1}^{\infty}\frac{x^{n}}{n!}=\frac{e^{x}-1}{x}$

   The $n$ cancels, and the remaining series is the exponential one minus its $n=0$ term. This is the move that makes the problem tractable.

3. 3

   Integrate back, using $S(0)=0$. Every term of $S$ vanishes at $x=0$, so

   $S(x)=\int_{0}^{x}\frac{e^{t}-1}{t}\,dt$

   The integrand is well behaved at $t=0$, where it tends to $1$, so there is no singularity to worry about.

4. 4

   Examine the integrand far to the left. For $t\to-\infty$ we have $e^{t}\to 0$, so

   $\frac{e^{t}-1}{t}\approx\frac{-1}{t}$

   which is positive for $t<0$ and decays only like $1/|t|$ — too slowly for the integral to settle.

5. 5

   Conclude that the integral diverges. Writing $S(x)=-\int_{x}^{0}\frac{e^{t}-1}{t}dt$ and comparing with $\int_{x}^{0}\frac{-1}{t}dt=\ln|x|\to+\infty$, the integral grows without bound, so

   $\lim_{x\to-\infty}S(x)=-\infty$

6. 6

   Confirm the logarithmic rate numerically. Summing the series directly: $S(-10)=-2.8798$, $S(-20)=-3.5729$, $S(-30)=-3.9784$. The asymptotic prediction $-\left(\ln|x|+\gamma\right)$ with $\gamma=0.5772$ gives $-3.9784$ at $x=-30$ ✓ — divergence, but only at logarithmic speed.