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$$\lim_{n\to \infty}\sum_{r=1}^n \tan^{-1} \dfrac{2r+1}{r^4+2r^3+r^2+1}$$

How am I supposed to do it? One thing I see here is

$$\lim_{n\to \infty}\sum_{r=1}^n \tan^{-1} \dfrac{2r+1}{(r^2+r)^2+1}$$

Here derivative of $r^2+r$ is $2r+1$. (if it helps)

It's final answer is

$\pi/4$

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  • $\begingroup$ What you're adding are a bunch of angles. Maybe draw the corresponding right triangles (putting the relevant angles next to one another) and come up with some argument from there? $\endgroup$ – Arthur Jan 30 '16 at 7:31
  • $\begingroup$ @Arthur After adding values of $r=1$ and $r=2$, I got $\tan^{-1}(3/5)+\tan^{-1}(5/37)$ = 0.67 radians = 38 degrees. But adding like that manually so many times won't make any sense. $\endgroup$ – manshu Jan 30 '16 at 7:42
  • $\begingroup$ Unless, if you draw the triangles, you can find some geometrical proof that the opposite leg never gets infinite, but diverges, for instance. I dunno, it was just an idea. $\endgroup$ – Arthur Jan 30 '16 at 7:46
  • $\begingroup$ @Arthur I don't think drawing would help. The answer is one algebraic trick away rather than in a geometric insight. $\endgroup$ – user230452 Jan 30 '16 at 8:18
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Hint

Since $$\dfrac{2r+1}{(r^2+r)^2+1}=\dfrac{(r+1)^2-r^2}{1+r^2(r+1)^2}$$ then $$\tan^{-1} \Big(\dfrac{2r+1}{(r^2+r)^2+1}\Big)= \tan^{-1}(1+r)^2- \tan^{-1}(r^2) $$ which beautifully telecospes.

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  • $\begingroup$ So is there any use of limit or is it just for a show coz I am getting the answer without any use of it. $\endgroup$ – manshu Jan 30 '16 at 7:52
  • $\begingroup$ It just telescopes !! $\endgroup$ – Claude Leibovici Jan 30 '16 at 7:53
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    $\begingroup$ @manshu It's not just for show. $\sum_{r = 1}^\infty$ is short-hand for $\lim_{n \to \infty}\sum_{r = 1}^n$, so the $\lim$ would actually be there anyway, just hidden away in notation. $\endgroup$ – Arthur Jan 30 '16 at 7:53
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You aren't too far from the answer. Credit on discovering what $(r+1)^2$ was. Write $r^2+r$ as $r(r+1)$. Squaring that would give you $r^2$ and $(r+1)^2$. And the difference of those two terms is the numerator. Now, it's in the form of $\tan(A-B)$.

Now, telescope this sum which becomes $\arctan(r)- \arctan(1)$

$\pi/2 - \pi/4 = \pi/4$

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  • $\begingroup$ thanks. but this is already covered in the answer given by claude. $\endgroup$ – manshu Jan 30 '16 at 8:28
  • $\begingroup$ @user230452. You are totally right. I was editing (typing the actuel first line) when your answer came (just check at what time I edited). $+1$ for your answer. Cheers. $\endgroup$ – Claude Leibovici Jan 30 '16 at 8:33
  • $\begingroup$ @ClaudeLeibovici Thanks. I up voted your answer too. $\endgroup$ – user230452 Jan 30 '16 at 8:52

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