Using

"assume(k, integer);
eval(sin(k*Pi));
                               0"

@nm 

It is an ODE with separated variables - almost a mental arithmetic task if you know some antiderivatives.

@janhardo 

The history of this problem can be traced back:
https://books.google.de/books?id=fmRkK_ho2vUC&printsec=frontcover&hl=de&source=gbs_ge_summary_r&cad=0#v=twopage&q&f=true
The theoretical background is also outlined on page 313 in Problem 171. But I was mainly interested in what Maple does with it. Now I have a lot to learn about it :-) It's impressive how the theory, which once seemed so dry, can now be experienced "alive and in color."

Many thanks.

@dharr 

...... is close to the theoretical result 5/2, which can be achieved with a trick using pen and paper.

@Kitonum 

...I had hoped that using gcd(m,n)=1 for the summation an index set of ordered pairs could be generated.

@Carl Love 

"Since the series is alternating, I decided to pair consecutive terms to make it a series of positive terms:" ...is only permitted if convergence is proven. However, this is clearly the case here according to the Leibnitz criterion for alternating series, and you have applied it.

@vv and @Carl Love

Both solutions have been instructive and enjoyable for me.

@janhardo 

...I would like to note:
In this original problem, I am solely interested in how a clever sequence of commands in Maple leads step by step from input to the result. A classic solution using "per pedes ;-)" calculations and the correct solution pi/4+1/2*ln(2) have long been known to me, and this is not the subject of my question. Since I am quite sensitive to convergence issues (especially regarding CAS), I missed some of the individual steps of your solution in your post. This is probably where the misunderstanding stems from—sorry. Therefore, a corresponding addition to your post would be helpful for my Maple learning success.

@janhardo 

......to put the factor j in front of the infinite sum over k?

@vv 

... and thank You.

@vv 

... the Riemann sum I know. As far as I remember, this problem comes from a short book by Williams and Hardy. But I'm interested in getting a result in Maple directly, without much preparation- just a maple exercise for me.

@dharr 

"0 means an exact result" That's exactly what I was asking for - thank you very much. (In return, I'll reveal solutions 11 and 7 ;-) ).

A possible arithmetic relationship between the numbers could be that the sum of two neighboring numbers in a row equals the number below it in the middle. Therefore, x=15968, y=15725, and z=47350 would be one solution. But it might not be the only one. Let's see what the prime factors say...

@salim-barzani 

Below, I would like to use the notation from your original post. Since it has now been established that f is a positive real number not equal to 1, as already mentioned, we have an ordinary differential equation. @janhardo cleverly reduced this to a Riccati-type differential equation using substitution and solved it.
The function U in your second post is very likely (I don't know the physical background) a linear combination of basis functions, which should be represented methodically more conveniently as exponential functions. This linear combination spans a finite-dimensional function space and is supposed to provide an approximate solution to another equation. In this specific case, the coefficients a must be determined according to criteria unknown to me. It is somewhat reminiscent of the Ritz/Galerkin method.
In any case, thanks to @janhardo, the function h has now been calculated in the exponent of f.

You write that f is an exponential function. However, the notation in equation (5) suggests that f is a number (regardless of whether it is real or complex). Or is f supposed to be a function of the variable phi from some arithmetic or other combination of the exponential function exp(phi)? This could easily be reformulated according to the power law. If f is a number, then equation (5) is a first-order ordinary differential equation with separated variables. Integrating the reciprocal of the right-hand side might then be somewhat challenging. But if f is meant to be a function f(phi), then a symbolic solution becomes very problematic.