def test_evalf_sum():
assert Sum(n, (n, 1, 2)).evalf() == 3.
assert Sum(n, (n, 1, 2)).doit().evalf() == 3.
# the next test should return instantly
assert Sum(1/n, (n, 1, 2)).evalf() == 1.5
# issue 8219
assert Sum(E/factorial(n), (n, 0, oo)).evalf() == (E*E).evalf()
# issue 8254
assert Sum(2**n*n/factorial(n), (n, 0, oo)).evalf() == (2*E*E).evalf()
# issue 8411
s = Sum(1/x**2, (x, 100, oo))
assert s.n() == s.doit().n()
def test_karr_convention():
# Test the Karr summation convention that we want to hold.
# See his paper "Summation in Finite Terms" for a detailed
# reasoning why we really want exactly this definition.
# The convention is described on page 309 and essentially
# in section 1.4, definition 3:
#
# \sum_{m <= i < n} f(i) 'has the obvious meaning' for m < n
# \sum_{m <= i < n} f(i) = 0 for m = n
# \sum_{m <= i < n} f(i) = - \sum_{n <= i < m} f(i) for m > n
#
# It is important to note that he defines all sums with
# the upper limit being *exclusive*.
# In contrast, diofant and the usual mathematical notation has:
#
# sum_{i = a}^b f(i) = f(a) + f(a+1) + ... + f(b-1) + f(b)
#
# with the upper limit *inclusive*. So translating between
# the two we find that:
#
# \sum_{m <= i < n} f(i) = \sum_{i = m}^{n-1} f(i)
#
# where we intentionally used two different ways to typeset the
# sum and its limits.
i = Symbol("i", integer=True)
k = Symbol("k", integer=True)
j = Symbol("j", integer=True)
# A simple example with a concrete summand and symbolic limits.
# The normal sum: m = k and n = k + j and therefore m < n:
m = k
n = k + j
a = m
b = n - 1
S1 = Sum(i**2, (i, a, b)).doit()
# The reversed sum: m = k + j and n = k and therefore m > n:
m = k + j
n = k
a = m
b = n - 1
S2 = Sum(i**2, (i, a, b)).doit()
assert simplify(S1 + S2) == 0
# Test the empty sum: m = k and n = k and therefore m = n:
m = k
n = k
a = m
b = n - 1
Sz = Sum(i**2, (i, a, b)).doit()
assert Sz == 0
# Another example this time with an unspecified summand and
# numeric limits. (We can not do both tests in the same example.)
f = Function("f")
# The normal sum with m < n:
m = 2
n = 11
a = m
b = n - 1
S1 = Sum(f(i), (i, a, b)).doit()
# The reversed sum with m > n:
m = 11
n = 2
a = m
b = n - 1
S2 = Sum(f(i), (i, a, b)).doit()
assert simplify(S1 + S2) == 0
# Test the empty sum with m = n:
m = 5
n = 5
a = m
b = n - 1
Sz = Sum(f(i), (i, a, b)).doit()
assert Sz == 0
e = Piecewise((exp(-i), Mod(i, 2) > 0), (0, True))
s = Sum(e, (i, 0, 11))
assert s.n(3) == s.doit().n(3)