@nm 

Reply to quotes:

"But without using the limit, how else will one find the constants of integration? Are you saying the series must be assumed to have  uniform convergence to use the limit as above?"

Yes.

"I assume Maple also used the limit internally to solve for C2 in this example and give the answer it did. How else could it have found the solution it did otherwise?"

That's probably the case. In such cases, a hint from the software would be helpful.

@nm 

...it should also be noted that uniform convergence of the three summand series must be assumed in the open interval x > 0. Only then can y(t) be assumed to converge uniformly in this interval. And only then is it permissible to calculate y(0) as the sum of the limits of the three summand series.

@nm 

...but that doesn't change the fact that when applying the explicit equation form y = f(t,y), as used in proofs of the existence and uniqueness of solutions, the right-hand side is not continuous/Lipschtz continuous at x = 0. Then Maple's error message should look different, or perhaps a note should appear indicating that a solution was only possible after continuing the solution to the edge of the domain of f(t,y).

Regarding your maple_sol solution, I'd like to know what Maple does with the natural logarithm it contains for x = 0.
The general solution may be correct; I haven't checked it. But classically, no special solution is possible for the initial value at x = 0.

@dharr 

You're right. 2^m is correct instead of 2^(m+1). Thanks for this hint and the tips with the table.

@dharr 

How can coordinates be retrieved from the plot of my "test" file, and how is a table of values ​​created, e.g., from {(m; y) } with y from term (1) from my "test" file for, e.g., m=2...17?

...to a problem that shouldn't be forgotten. There are solutions in the literature that are about 100 years old, and they're quite challenging in theory. However, with today's computational capabilities, it's now possible to come up with new proof ideas and practice mathematical reasoning. Perhaps I'll report on that later ;-) .
BTW: I chose this problem because I want to learn how to use Maple commands from the example solutions so that I can work on specific problems in a targeted manner later. I would be very happy to receive more examples with strongly asymmetrical contours, which do not have to be convex.

...it helped a lot.

@mmcdara 

...I expressed myself in a confusing way, for which I apologize. I only posed the original problem to learn how to handle constraints (here, m not equal to n) in Maple. In this problem, the inner sum over n and the outer sum over m must be calculated uniquely.

If the series elements are written as a matrix in the usual way, then, depending on the transposition, they must be added first column by column and then row by row, omitting the "forbidden" elements on the main diagonal. Rearranging the summation order and commuting the elements leads to different limit values, depending on the desired result. This is based, for example, on the so-called "Great Rearrangement Theorem" (Cauchy). It is one of the most powerful theorems in series theory. Accordingly, commutating is only permitted if absolute convergence exists. Your cited example of problem 21 is proof that reversing the order produces different results.

Therefore, the only possible limit for the original problem is pi^2/8 in terms of absolute value, which, depending on the transpose of the matrix, is either pi^2/8 or -pi^2/8 (the latter is my solution).
I hope our misunderstanding has been cleared up.

@acer 

The "remainder term" (3/4)/m^2 can be calculated using the inner sum (over n), which is initially finite up to N. It is calculated using the transformation sum(1/(m^2-n^2)) = -1/(2*m)*sum(1/(n-m)-1/(n+m)) = 1/(2*m)*(sum (k=1 to N+m)1/k - sum(k=1 to N-m)1/k) - 3/(4*m^2). The two sums in the parentheses are harmonic sums (sum of consecutive reciprocals). These are replaced in the usual way by the natural logarithm, the Euler constant, and O(1/n). After the limit N-->00, the difference in the sums in the parentheses becomes zero. The remainder term remains. A new limit is created.
Unfortunately, I can't yet represent this in Maple. I only have an old derive file. If you wish, I can put them into a text.

@mmcdara 

In the cited exercise 21, the divergence of the double series is not proven. It is only calculated that reversing the order of summation results in different results. Reversing the order of summation is permitted for finite sums. For double series whose convergence is to be investigated, this must first be proven.

@dharr mmcdara 7379

My question isn't about the solution, which I know. Rather, I'd like to learn how to insert constraints into the command chain in Maple, using this specific problem as an example. Such tasks also exist in optimization problems with constraints. For example, how do you do this:
Determine the max or min of a term under the condition Term(...)=0 or greater than ... or ... not equal to ...?

@Carl Love 

How can sin(k*x)*cos(x)^k be expanded into a series of sin and cos? Using convert?

@Carl Love 

I'm afraid ;-) you've found the same solution I achieved a long time ago, partly with the help of terrible formulas from Gradstein/Ryshik. Fortunately, transforming trigonometric terms and calculating improper integrals are now much easier to do on a computer. After all, it was my goal to find out whether symbolic calculations could make such problems easier to solve.

Would you please make your code available for practice purposes?

@nm 

Thanks, learned something new again. Then I want to calculate a table for a specific k-sequence, e.g., k = 10, 20, 30, ...100. How do I do that?

@nm 

MaplePrimes is finally working again. Thank you for the link to the book. I had a quick "browsing" look. I found the problem and its solution.
Summary:
The book contains a lot of interesting information, is aimed at interested beginners, and whets the curiosity for more in-depth content not covered in the book. Fortunately, there are classics such as Kamke, Walter, Pontrjagin, etc. Given the large number of example problems, it may happen that a typo goes unnoticed. So, I'll stick with the statement that your equation for the initial value y(0)=0 has no solution. In principle, you've already provided a proof by contradiction yourself. And theoretically, even for the initial value y(0)=1, the general solution only becomes the final solution when one considers the continuation of the solution from the open continuity domain of the "right-hand side" to the boundary. But that's splitting hairs and a different chapter of theory, and it's not worth putting a lot of effort into simple problems. The book authors should comment and the odetest would have to be examined on a case-by-case basis.