Can't solve Sum in sqrt

[lilithxxx]

>>> from sympy import *
>>> x=symbols('x')
>>> Sum(sqrt(x), (x,1,oo)).doit()
Sum(sqrt(x), (x, 1, oo))

where the answer should have been oo(related to #14410)

[lilithxxx]

This is the same for all powers of x which are decimals. For example, Sum(x**2.5, (x, 1, oo)) ,
Sum(x**0.5, (x, 1, oo)) they all won't work.

Even Sum(x**2.0, (x, 1, oo)) won't work while Sum(x**2, (x, 1, oo)) does.

[vishalg2235]

@jksuom @normalhuman here for any positive number of result the sum will be oo, even for negative number of result like in #14410 new cases should be added.

[dnabanita7]

Actually,here @lilithxxx says it correct.In the floating point integer,Sum doesn't work because of doit() While it works perfectly if we use evalf().So,I guess doit() module has to be checked.
Can I take up this issue? Ping @oscarbenjamin

[dnabanita7]

from sympy import *
x=symbols('x')
s=Sum(x**2.0,(x,1,oo)).evalf()
print(s)

The output is:
0.e+390

[dnabanita7]

doit() is in PyPi.I can't find any refrence of PyPi in here.

[jksuom]

Sum.doit is defined in summations.py. It seems that it could be improved to return oo for divergent series with positive terms like Sum(sqrt(n), (n, 1, oo)).

[dnabanita7]

Ok,I am checking it. Can you please tell me how to know which functions is in which file. I use git grep but I couldn't find it. I also don't know how to keyboard interrupt to check where the program halts. Thank you so much for helping.

…

On Wed 13 Mar, 2019, 1:35 PM Kalevi Suominen, ***@***.***> wrote: Sum.doit is defined in summations.py. It seems that it could be improved to return oo for divergent series with positive terms like Sum(sqrt(n), (n, 1, oo)). — You are receiving this because you commented. Reply to this email directly, view it on GitHub <#14579 (comment)>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AeGb9z868kl3X1zxJfieH3LStOT5F3TJks5vWLFggaJpZM4TCADV> .

[asmeurer]

If you use IPython/Jupyter Sum? will tell you in what file it is defined.

[anutosh491]

I'll take this up , seems approachable for a relatively new contributor like me . Thanks!