Final Evaluation Form N 2: Fill & Download for Free

This took me a while, because I took the wrong turn in which I used:[math]\displaystyle \dfrac{n^4+n^2–1}{n^4+n^2+1}=1-\dfrac{2}{n^4+n^2+1}\tag{1}[/math]Which is fine at first thought, I rewrote the requested series into the form:[math]\displaystyle e-3+\sum_{m=1}^{\infty} \dfrac{1}{(m+1)! m (m+1)}\tag{2}[/math]And tried to find the exponential generating function belonging to this series. I couldn't find an easy way to solve the associated differential equation.In arriving at (2) I used a couple of tricks which one should apply immediately to the requested sum, instead of splitting the summand like suggested in (1).In fact one can compute the partial sums:[math]\displaystyle \sum_{n=0}^{m} \dfrac{n^4+n^2–1}{n!(n^4+n^2+1)}\tag{3}[/math]As a first step one could write:[math]n^4+n^2+1=(n^2+n+1)(n^2-n+1)\tag*{}[/math]This tells us that we may write:[math]\displaystyle \dfrac{2n}{n^4+n^2+1} =\dfrac{1}{n^2-n+1}-\dfrac{1}{n^2+n+1}\tag{4}[/math]Furthermore if we change [math]n[/math][math] \to n+1[/math] in [math]n^2-n+1[/math], we obtain:[math](n+1)^2-(n+1)+1=n^2+n+1\tag*{}[/math]With:[math]T_n = \dfrac{1}{n^2-n+1}\tag*{}[/math]we found the following telescoping series, with the closed form solution:[math]\displaystyle \sum_{n=0}^m \dfrac{2n}{n^4+n^2+1} =\sum_{n=0}^m T_{n}-T_{n+1}=T_0-T_{m+1}=1-\dfrac{1}{m^2+m+1}\tag*{}[/math]The key idea is to use (4) and try to write (3) as a telescoping series as well.A term of the telescoping sum must look like:[math]\displaystyle \dfrac{f(n)}{(n^2-n+1)n!}\tag*{}[/math]where [math]f(n)[/math] is hopefully some simple polynomial expression.If we compute:[math]\displaystyle \dfrac{f(n)}{(n^2-n+1)n!}-\dfrac{f(n+1)}{(n^2+n+1)(n+1)!}\tag*{}[/math]and write the result as one fraction, we should end up with:[math]\displaystyle \dfrac{n^4+n^2–1}{n! (n^4+n^2+1)}\tag{5}[/math]If we want to arrive at [math]n^4[/math] the degree of [math]f(n)[/math] should be at least [math]2[/math][math][/math].Instead of trying a general polynomial we just try [math]f(n)=n^2[/math] and hope for the best. Write as one fraction:[math]\displaystyle \dfrac{n^2 (n+1)! (n^2+n+1) - (n+1)^2 [/math][math]n[/math][math]! (n^2-n+1)}{(n^4+n^2+1) [/math][math]n[/math][math]! (n+1)!}\tag*{}[/math]Factor [math](n+1)![/math] from the numerator and denominator:[math]\displaystyle \dfrac{(n+1)! \left(n^2(n^2+n+1) - (n+1) (n^2-n+1)\right)}{(n+1)! \left( (n^4+n^2+1)n!\right)}\tag*{}[/math]Cancel the common factor and finally check:[math]n^2(n^2+n+1)-(n+1)(n^2-n+1)=n^4+n^2–1\tag*{}[/math]which shows that this particular choice of [math]f(n)[/math] was spot on! A lucky guess perhaps.But that means our question can be answered. We find:[math]\displaystyle \sum_{n=0}^{m} \dfrac{n^4+n^2–1}{n!(n^4+n^2+1)}=\sum_{n=0}^m T_n-T_{n+1} \tag*{}[/math]with:[math]\displaystyle T_n=\dfrac{n^2}{(n^2-n+1)n!}\tag*{}[/math]and obtain (after simplification):[math]\displaystyle \sum_{n=0}^{m} \dfrac{n^4+n^2–1}{n!(n^4+n^2+1)}=T_0-T_{m+1}=-\dfrac{(m+1)}{(m^2+m+1)m!}\tag*{}[/math]The answer to the question is therefore: [math]0[/math].

I will use residues from Complex Analysis to compute the value of this series. The main idea is that for a convergent series of the form [math]\sum_{-\infty}^{\infty} f(n)[/math] whose summand [math]f(z)[/math] satisfies mild growth restrictions (as does the given summand in this problem), then the Residue Theorem under a suitable square contour yields[math]\sum_{-\infty}^{\infty} f(n) = -\sum (\text{residues of } \pi \cot(\pi z) f(z) \text{ at the poles of } f(z)).[/math]For this derivation, we initially assume that [math]k[/math] is a real constant not equal to 0.We first compute the residues of [math]\pi \cot(\pi z) \cdot \frac{1}{z^2 + k}[/math] at the poles of [math]f(z) = \frac{1}{z^2 + k}[/math].We see that the poles of [math]f(z)[/math] are given by [math]z = \pm \sqrt{-k}[/math]. Since both of these are simple poles, their respective residues for [math]\displaystyle \frac{\pi \cot(\pi z)}{z^2 + k}[/math] equal[math]\displaystyle \lim_{z \to \pm \sqrt{-k}} (z - (\pm \sqrt{-k})) \cdot \frac{\pi \cot(\pi z)}{z^2 + k} = \lim_{z \to \pm \sqrt{-k}} \frac{\pi \cot(\pi z)}{z \pm \sqrt{-k}} = \frac{\pi \cot(\pi \sqrt{-k})}{2\sqrt{-k}}. \tag*{}[/math]Therefore, we obtain[math]\displaystyle \sum_{-\infty}^{\infty} \frac{1}{n^2 + k} = -\Big(\frac{\pi \cot(\pi \sqrt{-k})}{2\sqrt{-k}} + \frac{\pi \cot(\pi \sqrt{-k})}{2\sqrt{-k}}\Big) = -\frac{\pi \cot(\pi \sqrt{-k})}{\sqrt{-k}}. \tag*{}[/math]However, by examining the index of summation and using symmetry, we have[math]\displaystyle \sum_{-\infty}^{\infty} \frac{1}{n^2 + k} = \frac{1}{k} + [/math][math]2[/math][math] \sum_{n=1}^{\infty} \frac{1}{n^2 + k}.\tag*{}[/math]Therefore, we obtain[math]\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2 + k} = -\frac{1}{2} \Big(\frac{\pi \cot(\pi \sqrt{-k})}{\sqrt{-k}} + \frac{1}{k}\Big). \tag*{}[/math]Now, we clean up this answer a little further.If [math]k > 0[/math], we can write [math]k = a^2[/math] for some [math]a > 0[/math], and using the fact that [math]\cot(ai) = -i\coth{a}[/math], our solution simplifies to[math]\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2 + a^2} = -\frac{1}{2} \Big(\frac{\pi \cot(a \pi i)}{ai} + \frac{1}{a^2}\Big) = \frac{1}{2a^2} \Big(a\pi \coth(a \pi) - 1\Big). \tag*{}[/math]If [math]k < 0[/math], we can write [math]k = -a^2[/math] for some [math]a > 0[/math]. One extra subtlety is that [math]a[/math] can not be a positive integer, because this forces one of the terms in the series to have division by 0.Assuming that [math]k = -a^2[/math] for some non-integer [math]a > 0[/math], our solution simplifies to[math]\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2 - a^2} = \frac{1}{2a^2} \Big(1 - \pi a \cot(\pi a)\Big). \tag*{}[/math]Finally to deal with the case [math]k = 0[/math], we can use the series where [math]k > 0[/math] and let [math]a \to 0^+[/math]:[math]\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \lim_{a \to 0^+} \frac{1}{2a^2} \Big(a\pi \coth(a \pi) - 1\Big) = \frac{\pi^2}{6}. \tag*{}[/math]

*A2AI’m going to demonstrate a way to evaluate this sum that doesn’t require much more than a year or two of calculus. I think Donald Hartig already showed the basic principles, but let me elaborate just a bit more.We wish to find[math]\displaystyle S = \sum_{n=1}^{\infty}{\dfrac{n^2}{2^n}}[/math]Let’s first consider a more familiar series[math]\displaystyle S^* = \sum_{n=1}^{\infty}{\dfrac{1}{2^n}}[/math]Most will quickly know the value of this sum as [math]S^* = 1[/math]. However, we can consider a more general form of this sum[math]\displaystyle S(x) = \sum_{n=1}^{\infty}{\dfrac{x^n}{2^n}}[/math]We can see that this sum, in general, evaluates to[math]S(x) = \dfrac{x/2}{1-x/2} = \dfrac{x}{2-x}[/math]What happens when we differentiate this? Well…[math]\begin{align*} S'(x) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \sum_{n=1}^{\infty}{\dfrac{x^n}{2^n}} &= \dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{x}{2-x} \\ &= \sum_{n=1}^{\infty}{\dfrac{nx^{n-1}}{2^n}} &= \dfrac{2}{(2-x)^2} \end{align*}[/math]Neat-o! We can first multiple by [math]x[/math] to get rid of that pesky [math]x^{n-1}[/math].[math]\displaystyle xS'(x) = \sum_{n=1}^{\infty}{\dfrac{nx^n}{2^n}} = \dfrac{2x}{(2-x)^2}[/math]Call this [math]T(x)[/math]. Differentiating this guy yields[math]\begin{align*} T'(x) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \sum_{n=1}^{\infty}{\dfrac{nx^n}{2^n}} &= \dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{2x}{(2-x)^2} \\ &= \sum_{n=1}^{\infty}{\dfrac{n^2x^{n-1}}{2^n}} &= \dfrac{2(2+x)}{(2-x)^3} \end{align*}[/math]How does this help? Well, look at [math]T'(1)[/math]:[math]T'(1) = \displaystyle \sum_{n=1}^{\infty}{\dfrac{n^2 1^{n-1}}{2^n}} = \sum_{n=1}^{\infty}{\dfrac{n^2}{2^n}}[/math]It’s our series! And look, we also have a formula for it![math]T'(1) = \dfrac{2(2+1)}{(2-1)^4} = \dfrac{2\times 3}{1} = 6[/math]And we have arrived at our final answer.