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I found this sum in the mathematial induction chapter of The art of Computer Programming and i have no idea how to solve it.

$\dfrac{1^3}{1^4+4}-\dfrac{3^3}{3^4+4} + ... +\dfrac{(-1)^n(2n+1)^3}{(2n+1)^4+4} $

I tried writing it as $\dfrac{1^3}{1*1^3+4}-\dfrac{3^3}{3*3^3+4} + ... +\dfrac{(-1)^n(2n+1)^3}{(2n+1)*(2n+1)^3+4} $

and then writing it as

$\dfrac{1}{1*1^3+4}-\dfrac{3+5}{3*3^3+4} +...+\dfrac{(-1)^n(((2n+1)^2-(2n+1)+1)+...+((2n+1)^2+(2n+1)-1))}{(2n+1)*(2n+1)^3+4} $

but did not know how to continue.

I also tried writing it as $\dfrac{1}{1+\dfrac{4}{1^3}}-\dfrac{1}{3+\dfrac{4}{3^3}} + ... +\dfrac{(-1)^n}{2n+1+\dfrac{4}{(2n+1)^3}} $ but without succes.

Did
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Andrei Mihailescu
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  • Where are you stuck? How far did you get in attempting to answer it? (Note, I'm assuming you put effort into the problem before posting here, else, if you've not yet done anything, please know that this site is not a "do all the work for me" site.) – amWhy Nov 03 '17 at 17:54
  • I tried a few things ( like writing x^4 as 3*x^3 and writing the cubes as a sum of odd numbers or dividing by x^3 ) but nothing worked . – Andrei Mihailescu Nov 03 '17 at 17:58
  • Could you please point out the approach you used ?Thank you! – Andrei Mihailescu Nov 03 '17 at 17:59
  • Andrei It would be great if you included your work, correct or incorrect (in an edit to your post). It simply helps us to know your efforts, and if there are mistakes, it helps us hone our answer to your needs. – amWhy Nov 03 '17 at 18:01
  • I described my attempts. – Andrei Mihailescu Nov 03 '17 at 18:34
  • Thanks, @Andrei. If your question gets put on hold, ping me, and I'll vote to reopen. I really appreciate your effort here! (When I say "ping", I mean make a comment using @amWhy – amWhy Nov 03 '17 at 18:36

2 Answers2

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Keyword: Concatenation.

First note that, for every $x$, $$\frac{4x^3}{x^4+4}=\frac{4x^3}{(x^2+2)^2-4x^2}=\frac{x^2}{(x-1)^2+1}-\frac{x^2}{(x+1)^2+1}$$ hence the partial sum $S_{2N+1}$ of $2N+1$ terms is such that $$4S_{2N+1}=\frac45+\sum_{n=1}^N\left(\frac{4(4n+1)^3}{(4n+1)^4+4}-\frac{4(4n-1)^3}{(4n-1)^4+4}\right)$$ that is, $$4S_{2N+1}=\frac45+U_N-V_N$$ where $$U_N=\sum_{n=1}^N\frac{(4n+1)^2}{(4n)^2+1}+\frac{(4n-1)^2}{(4n)^2+1}=\sum_{n=1}^N\frac{2((4n)^2+1)}{(4n)^2+1}=2N$$ and $$V_N=\sum_{n=1}^N\frac{(4n+1)^2}{(4n+2)^2+1}+\frac{(4n-1)^2}{(4n-2)^2+1}$$ Thus, $$V_N=\frac95-\frac{(4N+3)^2}{(4N+2)^2+1}+\sum_{n=1}^N\frac{(4n+1)^2}{(4n+2)^2+1}+\frac{(4n+3)^2}{(4n+2)^2+1}$$ that is, $$V_N=\frac95-\frac{(4N+3)^2}{(4N+2)^2+1}+\sum_{n=1}^N\frac{2((4n+2)^2+1)}{(4n+2)^2+1}=\frac95-\frac{(4N+3)^2}{(4N+2)^2+1}+2N$$ Coming back to $S_{2N+1}$, one gets $$4S_{2N+1}=\frac45-\frac95+\frac{(4N+3)^2}{(4N+2)^2+1}=\frac{8N+4}{(4N+2)^2+1}$$ hence, finally,

$$S_{2N+1}=\frac{2N+1}{4(2N+1)^2+1}$$

The even numbered sums $S_{2N}$ are solvable by a similar treatment.

Fun fact: $$\lim_{N\to\infty}S_N=0$$

Did
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0

The exercise asks you to find and prove a closed form. You can take a guided approach (e.g. Gosper's algorithm), but the intention is probably that you work out a few cases by hand, guess the desired form, and then prove it.

$$\begin{eqnarray} n=1 & \implies & \frac{1}{5} \\ n=2 & \implies & \frac{1}{5} - \frac{27}{85} = \frac{-50}{425} = \frac{-2}{17} \\ n=3 & \implies & \frac{-50}{425} + \frac{125}{629} = \frac{21675}{267325} = \frac{3}{37} \\ n=4 & \implies & \frac{21675}{267325} - \frac{343}{2405} = \frac{-39564100}{642916625} = \frac{-4}{65} \\ \end{eqnarray}$$

So it's worth hazarding a guess that the desired form is $\frac{(-1)^{n+1}n}{p(n)}$. At this point you could break out the algebra to see what $p(n)$ would work, or you could guess (or know, if you understand the theory behind Gosper's algorithm) that $p(n)$, if it has a closed form, must be a polynomial and see whether the cubic fit through $(1,5), (2,17), (3,37), (4,65)$ works. If not, evaluate another term and fit a quartic...

Peter Taylor
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