Find the sum of the following infinite series
$$\frac {1}{3!} + \frac {4}{4!} + \frac {9}{5!} + \ldots$$
I was trying this question ,but I couldn't get it. This is homework that my teacher has given me. I solved this question using d'Alembert's ratio test but I didn't get a finite value.
If anyone could help me, that would be much appreciated. Thank you
It looks like you're trying to calculate
$$\sum_{n=1}^{\infty}\frac{n^{2}}{(n+2)!}$$
If you've covered Taylor series, they will be one way to solve this problem.
The Taylor series for $e^{x}$ about $x=0$ is:
$$\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$$
Now let's truncate the first three terms then divide the whole thing by $x^{2}$, so the first term is now $\frac{x}{3!}$. So we have:
$$\frac{e^{x}-1-x-\frac{x^{2}}{2}}{x^{2}} = \sum_{n=3}^{\infty}\frac{x^{n-2}}{n!} = \sum_{n=1}^{\infty}\frac{x^{n}}{(n+2)!} =\frac{x^{1}}{3!}+\frac{x^{2}}{4!}+\frac{x^{3}}{5!}+\ldots$$
This is almost what we want, if we set $x=1$. We just need to introduce a factor of $n^{2}$ to each term. Can you do that?
Hint:
$\frac{d}{dx}(x^{n}) = nx^{n-1}$, so $\frac{d}{dx}\left(x\frac{d}{dx}(x^{n})\right) =\,?$
It seems that you want to compute $$\sum_{n=0}^\infty\dfrac{n^2}{(n+2)!}$$ So, let us consider $$y=\sum_{n=0}^\infty\dfrac{n^2}{(n+2)!}x^{n+2}$$ and differentiate twice to get $$y''=\sum_{n=0}^\infty\dfrac{n^2}{n!}x^{n}=\sum_{n=0}^\infty\dfrac{n(n-1)+n}{n!}x^{n}=\sum_{n=0}^\infty\dfrac{n(n-1)}{n!}x^{n}+\sum_{n=0}^\infty\dfrac{n}{n!}x^{n}$$ that is to say $$y''=x^2\sum_{n=0}^\infty\dfrac{n(n-1)}{n!}x^{n-2}+x\sum_{n=0}^\infty\dfrac{n}{n!}x^{n-1}=x^2\left(\sum_{n=0}^\infty\dfrac{x^n}{n!} \right)''+x \left(\sum_{n=0}^\infty\dfrac{x^n}{n!} \right)'$$ That is to say $$y''=x^2e^x+x e^x$$ Using integration by parts, you should arrive to $$y'=e^x \left(x^2-x+1\right)+c_1$$ $$y=e^x \left(x^2-3 x+4\right)+c_1x+c_2$$ But, $$x=0\implies y=0\implies 4+c_2=0\implies c_2=-4$$ $$x=0\implies y'=0\implies 1+c_1=0\implies c_1=-1$$ All of the above then makes $$y=e^x \left(x^2-3 x+4\right)-x-4$$ Now, making $x=1$ leads to $$\sum_{n=0}^\infty\dfrac{n^2}{(n+2)!}=2e-5$$
By elementary means:
$$n^2=(n+2)(n+1)-3(n+2)+4$$ and
$$\sum_{n=0}^\infty\frac{(n+2)(n+1)-3(n+2)+4}{(n+2)!}=\sum_{n=0}^\infty\frac1{n!}-3\sum_{n=0}^\infty\frac1{{(n+1)}!}+4\sum_{n=0}^\infty\frac1{{(n+2)}!}.$$
Hence the sum is
$$e-3(e-1)+4(e-1-1).$$
The method extends to any polynomial divided by a factorial, giving a sum of the form $ae+b$ where $a,b$ are rational linear combinations of the polynomial coefficients (Stirling numbers of the second kind involved).
