Jtdylktuy

I need to find a sum and number of terms in an infinite series if ε ∈ (0;1) and x ∈ (1;5):

$$sum_{k=0}^∞ frac{k^2x^k}{(k+1)!}$$

I was able to convert it into a simpler form, but stuck there:

$$lim_{k→∞} frac{x^k×k^2}{k!×(k+1)} = e^x frac{k^2}{k+1}$$

Can you suggest a direction to simplify the $frac{k^2}{k+1}$ portion, please?

$$
begin{align}
sum_{k=0}^inftyfrac{k^2x^k}{(k+1)!}
&=sum_{k=0}^inftyfrac{((k+1)k-(k+1)+1),x^k}{(k+1)!}tag1\
&=sum_{k=1}^inftyfrac{x^k}{(k-1)!}-sum_{k=0}^inftyfrac{x^k}{k!}+sum_{k=0}^inftyfrac{x^k}{(k+1)!}tag2\
&=xsum_{k=0}^inftyfrac{x^k}{k!}-sum_{k=0}^inftyfrac{x^k}{k!}+frac1xleft(sum_{k=0}^inftyfrac{x^k}{k!}-1right)tag3\
&=(x-1)e^x+frac{e^x-1}xtag4\[3pt]
&=frac{(x^3+1)e^x-(x+1)}{x(x+1)}tag5
end{align}
$$
Explanation:
$(1)$: $k^2=(k+1)k-(k+1)+1$
$(2)$: cancel numerators and denominators
$phantom{text{(2):}}$ since $(k+1)k=0$ when $k=0$, we can remove the $k=0$ term from the first sum
$(3)$: substitute $kmapsto k+1$ in the first sum and $kmapsto k-1$ in the last sum
$(4)$: recognize the series for $e^x$
$(5)$: simplify the fraction

$$sum_{k=0}^∞ frac{k^2x^k}{(k+1)!}{\=sum_{k=0}^∞ frac{(k^2+k)x^k}{(k+1)!}-sum_{k=0}^∞ frac{kx^k}{(k+1)!}\=sum_{k=1}^∞ frac{x^k}{(k-1)!}-sum_{k=0}^∞ frac{kx^k}{(k+1)!}\=xe^x-sum_{k=0}^∞ frac{kx^k}{(k+1)!}}$$also $$sum_{k=0}^∞ frac{kx^k}{(k+1)!}=sum_{k=0}^∞ frac{x^k}{k!}-sum_{k=0}^∞ frac{x^k}{(k+1)!}=e^x-sum_{k=0}^∞ frac{x^{k+1}}{x(k+1)!}=e^x-{1over x}(e^x-1)$$therefore $$sum_{k=0}^∞ frac{k^2x^k}{(k+1)!}=e^x(x-1+{1over x})-{1over x}$$

I had very serious mistakes in my answer. Trying to fix them now.

After Yuriy S's comment, if you need to find $k$ such that
$$frac{k^2, x^k}{(k+1)!}< 10^{-d} tag 1$$
Since $$frac{k^2}{(k+1)!}=frac k{k+1} frac 1 {(k-1)!}sim frac 1 {(k-1)!}$$
we can approximate $(1)$ by equation
$$(k-1)! = ( x,{10^d}),x^{k-1}$$ or, simpler,
$$n!= (x,10^d), x^n qquad text{where }qquad n=k-1$$ If you look at this question of mine, you will see a magnificent approximation of the inverse factorial function which was proposed by @robjohn.

Applied to this case, this would give
$$color{blue} { ksim e, x,e^{W(t)}+frac 12} qquad text{where }qquad color{blue} {t=frac {log left(frac{x, 10^{2 d}}{2 pi }right) } {2e x }}tag 2$$ where appears $W(.)$ which is Lambert function.

Let us try for a few values of $x$ and $d$. The next table reports in the real domain the value given by the approximation $(2)$ as well as the exact solution of $(1)$. You just need to use the next integer.
$$left(
begin{array}{ccc}
x & d & text{approximation} & text{exact} \
1 & 4 & 8.33367 & 8.28080 \
1 & 5 & 9.41743 & 9.37320 \
1 & 6 & 10.4438 & 10.4056 \
1 & 7 & 11.4258 & 11.3922 \
1 & 8 & 12.3721 & 12.3420 \
1 & 9 & 13.2888 & 13.2615 \
& & & \
2 & 4 & 12.0127 & 11.9689 \
2 & 5 & 13.2892 & 13.2517 \
2 & 6 & 14.5002 & 14.4674 \
2 & 7 & 15.6595 & 15.6302 \
2 & 8 & 16.7765 & 16.7500 \
2 & 9 & 17.8579 & 17.8338 \
& & & \
3 & 4 & 15.3123 & 15.2744 \
3 & 5 & 16.7140 & 16.6810 \
3 & 6 & 18.0471 & 18.0178 \
3 & 7 & 19.3250 & 19.2987 \
3 & 8 & 20.5572 & 20.5333 \
3 & 9 & 21.7508 & 21.7288 \
& & & \
4 & 4 & 18.4409 & 18.4071 \
4 & 5 & 19.9348 & 19.9051 \
4 & 6 & 21.3592 & 21.3326 \
4 & 7 & 22.7268 & 22.7028 \
4 & 8 & 24.0470 & 24.0250 \
4 & 9 & 25.3267 & 25.3063 \
& & & \
5 & 4 & 21.4717 & 21.4412 \
5 & 5 & 23.0379 & 23.0108 \
5 & 6 & 24.5347 & 24.5104 \
5 & 7 & 25.9744 & 25.9521 \
5 & 8 & 27.3656 & 27.3452 \
5 & 9 & 28.7152 & 28.6963
end{array}
right)$$