May I know the answer for this with respect to

[dimension10]

May I know the answer for this with respect to t:

$$\sum_{n=1}^{\infty}(\frac{t}{2n}-\arctan(\frac{2t}{4n+1}))$$

Thanks in advance.

 

[hunt_mat]

Can I just expand the series as:
$$\frac{t}{2}\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\tan^{-1}\frac{2t}{4n+1}$$
and the first term is infinite unless t=0.

 

[micromass]

Can I just expand the series as:
$$\frac{t}{2}\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\tan^{-1}\frac{2t}{4n+1}$$
and the first term is infinite unless t=0.

No, that won't work. A difference of two divergent series may be divergent. A simple example is

$$\sum_{n=1}^{+\infty}{\frac{1}{n}-\frac{1}{n}}$$

I can show that the series of the OP converges using the integral test. But it seems that the OP wants the exact value of the series. I have no clue how to calculate that...

 

[Unit]

I have an idea, but I don't know where to go with it yet; try finding an expression whose derivative or integral is t/2n - arctan(2t/(4n+1)) and make that a function of t, and try to solve with power series manipulations.

 

[disregardthat]

No, that won't work. A difference of two divergent series may be divergent. A simple example is

There is no such thing as a difference between two divergent series. (sum 1/n) - (sum 1/n) and sum (1/n -1/n) are two very different things (the first one does not make sense at all).

The equivalence (sum a_n) + (sum b_n) = sum (a_n+b_n) is true (and makes sense) if and only if the sums exists (that is, converges).

 

[dimension10]

Using a calculator, I got infinity

 

[Donsig]

Writing t/(2n)-arctan(2*t/(4n+1)) as a series in t and then summing each coefficient as a series in n (in Maple) gives
(2-3/2*ln(2)-1/4*Pi)*t-1/48*Psi(2,5/4)*t^3+1/3840*Psi(4,5/4)*t^5-1/645120*Psi(6,5/4)*t^7+1/185794560*Psi(8,5/4)*t^9+O(t^11)
Psi(n,x) is the nth derivative of Psi(x), so it seems possible that the series may be expressible in terms of Psi(x). At least, this method might give you a reasonable approximation to the original series. I haven't thought carefully about the convergence issues, but I believe the original series does converge.

 

[dimension10]

Writing t/(2n)-arctan(2*t/(4n+1)) as a series in t and then summing each coefficient as a series in n (in Maple) gives
(2-3/2*ln(2)-1/4*Pi)*t-1/48*Psi(2,5/4)*t^3+1/3840*Psi(4,5/4)*t^5-1/645120*Psi(6,5/4)*t^7+1/185794560*Psi(8,5/4)*t^9+O(t^11)
Psi(n,x) is the nth derivative of Psi(x), so it seems possible that the series may be expressible in terms of Psi(x). At least, this method might give you a reasonable approximation to the original series. I haven't thought carefully about the convergence issues, but I believe the original series does converge.

t is a constant.

Anyway, I realized it should converge. It is related to the Riemann-Siegel Theta function.

$$\theta (t) = - \frac{t\, \gamma + t\;\mbox{ln}\: \pi}{2}-\arctan 2t + \sum_{n=1}^{\infty}(\frac{t}{2n}-\arctan\frac{2t}{4n+1})$$

and this theta function does converge, at least when the Z-function is concerned.

 

[bpet]

t is a constant.

Anyway, I realized it should converge. It is related to the Riemann-Siegel Theta function.

$$\theta (t) = - \frac{t\, \gamma + t\;\mbox{ln}\: \pi}{2}-\arctan 2t + \sum_{n=1}^{\infty}(\frac{t}{2n}-\arctan\frac{2t}{4n+1})$$

and this theta function does converge, at least when the Z-function is concerned.

Put

$$\arctan\frac{2t}{4n+1} = \int_0^t \frac{2(4n+1)}{(4n+1)^2+4x^2}dx$$

and interchanging the sum and integral and using Mathematica gives

$$\int_0^t \sum_{n=1}^{\infty} \frac{1}{2n} -\frac{2(4n+1)}{(4n+1)^2+4x^2} dx<br /> = \frac{\gamma t}{2} + \frac{1}{2i}(\log\Gamma(5/4+it/2)-\log\Gamma(5/4-it/2))$$

It should be possible to put the solution in real form, for example with the right integral representation of log(Gamma).