Find the sum of arctan(1/2n^2)

[Yegor]

$$\sum_{i=1}^{n} \arctan(\frac{1}{2n^2})$$

I have to find sum of first n terms.
I have no idea what to do with these arctan. Give me please any hint.

 

[dextercioby]

The answer is immediate.

$$\sum_{i=1}^{n} \arctan\left(\frac{1}{2n^{2}}\right) =n\arctan\left(\frac{1}{2n^{2}}\right)$$

Daniel.

 

[Yegor]

I got that
$$S_n=\arctan\frac{\tan(S_{n-1})+1/2n^2}{1-\frac{\tan(S_{n-1})}{2n^2}}$$
$$S_n$$ is sum of first n terms.
But I'm not sure if it could help.
I just started studying series by myself. Could you elaborate your solution, Dexter. I don't understand

 

[dextercioby]

I've already given you the answer.Yours is not the answer to the problem written in post #1.

Daniel.

 

[Yegor]

Oh, yes. It's my mistake.
$$\sum_{n=1}^{\infty} \arctan(\frac{1}{2n^2})$$ is supposed to be. Sorry

 

[dextercioby]

I haven't got a clue so far to prove that

$$\sum_{n=1}^{\infty} \arctan\left(\frac{1}{2n^{2}}\right) =\frac{\pi}{4}$$

Daniel.

 

[Dr.Brain]

I haven't got a clue so far to prove that

$$\sum_{n=1}^{\infty} \arctan\left(\frac{1}{2n^{2}}\right) =\frac{\pi}{4}$$

Daniel.

Integration Daniel?

 

[dextercioby]

Integrate what...?I don't see a pattern to form a Riemann sum.

Daniel.

 

[Dr.Brain]

Integrate what...?I don't see a pattern to form a Riemann sum.

Daniel.

I tried it out, no light.Where did you get thid problem from?

 

[Hurkyl]

Wee, this is a variant of my favorite problem! The way I know how to do it involves the application of a trig identity...

 

[Yegor]

Hm. Unfortunately it's too subtle hint for me.

 

[himanshu121]

$$\frac{1}{2n^2} = \frac{2}{4n^2}$$ u can use this
to get
$$arctan(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$$
You can use the identity for inverse
and try to figure out rest

 

[himanshu121]

ok just saw ur post::
arctana+arctanb = arctan(a+b)/(1-ab)

so u will have arctan{1/2²}= arctan(2n+1)-arctan(2n-1)

now u can substitute for n=1,2,3,...k u will finally have

arctan{1/n²}= arctan(2k+1)-arctan1
now as k=> inf u will have pi/2-pi/4 = pi/4


"what is here may be elsewhere, what is not here is nowhere."
Prayers are empowered by a clear mind and heart, and purity of intention. With clarity, our consciousness aligns with higher consciousness, where the forces of nature and laws of creation operate according to divine order

 

[Hurkyl]

Well, I was trying not to actually do his homework for him. :tongue2:

 

[PhilG]

At any rate, it is a very neat problem. I had not seen it before. Thanks to the OP for posting it. I wonder if there are other ways to do it ...?

 

[Yegor]

$$\arctan{1/2n^2}= \arctan(2n+1)-\arctan(2n-1)$$
Yes, it's very tricky!
I also got that $$S_n=\arctan(\frac{n}{n+1})$$ (This was my question)
This also leads to pi/4 when n->infinity.
Thank you very much to all.

 

[saltydog]

Thanks guys. Very clever. This is a summary which I take no credit for:

Using Himanshu's technique above:

$$\sum_{n=1}^{\infty}Arctan(\frac{1}{2n^2})$$

Let:

$$\frac{1}{2n^2}=\frac{2}{4n^2}=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$$

Now,

$$Arctan(a)-Arctan(b)=Arctan\left[\frac{a-b}{1+ab}\right]$$

Letting:

a=(2n+1)
b=(2n-1)

We have:

$$Arctan[\frac{1}{2n^2}]=Arctan\left[\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right]=Arctan(2n+1)-Arctan(2n-1)$$

Thus we have:

$$\sum_{n=1}^{\infty}Arctan(\frac{1}{2n^2})=\sum_{n=1}^{\infty}\left[Arctan(2n+1)-Arctan(2n-1)\right]$$

So that using a(x) and a(y) to save space we have:

$$\sum_{n=1}^{\infty}\left[Arctan(2n+1)-Arctan(2n-1)\right]=$$

$$(a3-a1)+(a5-a3)+(a7-a5)+(a9-a7)+ . . .+[a(2n+1)-a(2n-1)]+ . . .$$

Noting that all but -a1 and the a(2n+1) terms cancel, we take the limit of this sum and find we are left with:

$$\sum_{n=1}^{\infty}\left[Arctan(2n+1)-Arctan(2n-1)\right]\to a(\infty)-a(1)$$

Since:

$$Arctan(1)=\frac{\pi}{4}$$

and:

$$\mathop\lim\limits_{n\to\infty}Arctan(n)=\frac{\pi}{2}$$

we finally obtain:

$$\sum_{n=1}^{\infty}Arctan(\frac{1}{2n^2})=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$

That is beautiful!

 

[saltydog]

Oh yea, I almost forgot. What about these:

$$\sum_{n=1}^{\infty}ArcSin[\frac{1}{2n^2}]$$

$$\sum_{n=1}^{\infty}ArcCos[\frac{1}{2n^2}]$$

$$\sum_{n=1}^{\infty}ArcCot[\frac{1}{2n^2}]$$

I assume there must be an equivalent relation for the ArcCot. I'll look into it.
Wait a minute, isn't the cot=1/tan so can I propose that last one is just $\frac{4}{\pi}$?

 

[Hurkyl]

A sum of reciprocals is generally not the reciprocal of the sum.

One of the advantages of the tangent function is that its sum formula involves only tangents, so the corresponding formula for arctangents is nicer.

You could try constructing similar things for the others, I suppose. Start with a telescoping series, then apply an addition formula and see what you get.

 

[saltydog]

A sum of reciprocals is generally not the reciprocal of the sum.

One of the advantages of the tangent function is that its sum formula involves only tangents, so the corresponding formula for arctangents is nicer.

You could try constructing similar things for the others, I suppose. Start with a telescoping series, then apply an addition formula and see what you get.

Hello Hurkyl. You got to it before I could make amends. Yea, ArcCot and ArcCos will not converge. However, I suspect the ArcSin will.

You know, it's interesting just looking at the functions. Note that ArcTan and ArcSin are 0 at 0 where the other 2 are >0. Thus intutitively one would suspect taking an infinite sum of a function closer and closer to x=0 for functions which are >0 would not converge whereas if the functions are 0 at x=0, the sum I suspect could converge. Suppose that's all embodied in the various tests for convergence.

 

[saltydog]

I just seem to dig myself a hole in here sometimes. Suppose it's because my purpose for being here is to learn math as opposed to trying to show off. Anyway, concerning ArcSin. Quickest way for me to determine convergence is the integral test and yes I'm lazy and just used Mathematica but only because I've already done a bunch by hand. Anyway, the integral converges. So the limit of the sum exists. Now as far as what it is . . .

 

[PhilG]

I was wondering if it is possible to use complex variables to solve the original problem with arctan. I've seen some trick where you convert an infinite sum into a contour integral, but I don't remember exactly how it goes. I think it's called a Sommerfeld-Watson transformation, and you can use it to get the result $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$. Would that also work here?

 

[Hurkyl]

I have, in fact, seen someone solve a problem very similar to this one using complex analysis... so I'd imagine this one can be done as well. I don't remember the methodology, though.

 

[PhilG]

Hurkyl:

I googled Sommerfeld-Watson transform to find out more about it. Here's an example. If f(z) goes to zero faster than 1/z as z --> infinity, then the sum $\sum_{n=-\infty}^\infty f(n)$ can be found by computing the integral

$$\int_{\Gamma} f(z) \pi \cot \pi z \ dz = 2\pi i (\sum_{n=-N}^N f(n) \ + \ \sum_k \pi \cot \pi z_k * (\mbox{Res of f(z) at } z_k))$$

where, the integral is taken over the square of side 2N+1, centered at the origin. The integral goes to zero as N goes to infinity, and that allows you to compute the infinite series by summing a few residues.

Now it seems to me that arctan(1/2z^2) goes to zero like 1/2z^2 as z goes to infinity, so this method might work. However, doesn't arctan have a logarithmic branch point at zero? That might be annoying. And also, does arctan have any poles in the complex plane? I have forgotten how to determine this. Need to do some serious review.

Of course, I realize that the right way to do the problem is with the trig identity. I'm just curious to see if this way works too.

edit: as written, the above formula only makes sense if f(z) has simple poles.

 

[saltydog]

Of course, I realize that the right way to do the problem is with the trig identity.

"There were two paths in the forest.
I took the one less traveled and it has made all the difference"

Who said that?