# May I know the answer for this with respect to

• dimension10
In summary, the series \sum_{n=1}^{\infty}(\frac{t}{2n}-\arctan(\frac{2t}{4n+1})) converges and is related to the Riemann-Siegel Theta function. It can be expressed in terms of the Psi and Gamma functions, and can potentially be written in real form using the integral representation of log(Gamma). Further analysis and calculations are needed to determine the exact value of the series.
dimension10
May I know the answer for this with respect to t:

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

Last edited:

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.

hunt_mat said:
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...

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.

micromass said:
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).

Using a calculator, I got infinity

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.

Donsig said:
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.

dimension10 said:
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

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

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

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