Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.
N[Sum[1/(x^2 + 1), {x, 1, Infinity}], 5]
N[Sum[1/(x^2 + x + 1), {x, 1, Infinity}], 5]

1.0767

0.79815 + 0.*10^-6 I

What causes the strange number?

share|improve this question
5  
Note that your strange number $0. \times 10^{-6} \mathbf{i}$ is, within the precision you specified, equal to 0. –  Hurkyl Dec 22 '12 at 17:25

2 Answers 2

up vote 12 down vote accepted

Note that Mathematica can do the symbolic sums:

Sum[1/(x^2 + 1), {x, 1, Infinity}]

gives

$$ \frac{1}{2} (-1 + \pi \coth[\pi]) $$

and

Sum[1/(x^2 + x + 1), {x, 1, Infinity}]

gives

$$ -\frac{i \left(\psi ^{(0)}\left(1+\sqrt[3]{-1}\right)-\psi ^{(0)}\left(1-(-1)^{2/3}\right)\right)}{\sqrt{3}} $$ (where $\psi^{(0)}(x)$ is PolyGamma[0,x], and in fact if you apply FullSimplify this can be written in terms of HarmonicNumber as well). But you are asking for numeric versions of these to a specific number of significant figures. To that precision, the imaginary parts of the latter function do not quite cancel, and hence you get a small imaginary part.

Note also that using NSum[...] instead of N[Sum[...]] works better here, since it actually does the numerical sum by adding the terms, as opposed to calculating the symbolic form and then numerically evaluating it.

share|improve this answer
1  
(+1) but one note for completeness: in cases where Sum can't solve the sum symbolically (not applicable here), N[Sum[...] seems to be equivalent to NSum[...] according to the docs for NSum. –  Jens Dec 22 '12 at 17:31
    
Is there any reason that PolyGamma is used for this number as opposed to the first number? –  Fred Kline Dec 22 '12 at 18:20
    
Infinite sums are like integrals: even seemingly simple expressions can have complicated answers, and similar expressions can have very different answers. –  Andrew Jaffe Dec 22 '12 at 20:08
    
@FredKline This question mathematica.stackexchange.com/questions/17809/… discusses a related issue in Mathematica 8 and 9. –  Artes Feb 4 '13 at 20:04

There are several methods in Sum which may help in various cases. To find your second sum symbolically in apparently real form you can try, e.g.

Sum[1/(1 + x + x^2), {x, 1, Infinity}, Method -> "HypergeometricTermPFQ"]

enter image description here

However there are still other ways to get the same result without methods specified, e.g. :

s = Sum[1/(1 + x + x^2), {x, 1, Infinity}];

FullSimplify[ ComplexExpand //@ s ]

or

FullSimplify @ ExpToTrig @ s

Note that we had to MapAll (shorthand //@) ComplexExpand i.e. apply the latter to every subexpression of s

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.