Mathematica thinks (-1)^n is non-real

Question

When I ask Mathematica to evaluate

NSum[((-1)^n)/n, {n, 1, 100}]

it returns -0.688172 + 2.11297*10^-16 i

Why is this? (-1)^n is either 1 or -1. I don't know why it is returning a non-real number. Is there a way to avoid this?

Answer 1

Machine floating point arithmetic can produce strange results when the numbers get very small. Here are same ways to deal with it.

Suppress insignificant imaginary parts.

NSum[((-1)^n)/n, {n, 1, 100}] // Chop

-0.688172

Work with exact numbers until the very end (slower).

Sum[((-1)^n)/n, {n, 1, 100}] // N

-0.688172

Use Mathematica's arbitrary precision arithmetic rather than machine arithmetic.

NSum[((-1)^n)/n, {n, 1, 100}, WorkingPrecision -> 16]

-0.688172179310

Use the special method that NSum has for maintaining precision with series that have terms with alternating signs. This is likely to be the best solution for this particular sum.

NSum[((-1)^n)/n, {n, 1, 100}, Method -> "AlternatingSigns"]

-0.688172

Answer 2

NSum uses a combination of analytic and numeric methods. For instance, it computes

(-1)^102.
(*  1. - 1.95968*10^-14 I  *)

with a real exponent instead of an exact integer. It's the same as Exp[102. Log[-1]], which is the same as Exp[I * 102. * Pi], for that is how Power is computed when the exponent is Real. Rounding error in 102. * Pi introduces a small, complex part in the final result.

For more, see the output of

Trace[NSum[((-1)^n)/n, {n, 1, 100}]]

As @kglr pointed out, the way to make Mathematica handle the -1 power appropriately is with

NSum[((-1)^n)/n, {n, 1, 100}, Method -> "AlternatingSigns"]

The fact that Power with a negative base is going to be a complex-valued function when the exponent is an approximate Real can be countered, given the terms are real, by using Re:

NSum[Re[((-1)^n)/n], {n, 1, 100}]
(*  -0.688172  *)

A different approach that uses exact exponents and Real base:

Total[(-1.)^#/# &@Range[100]]
(*  -0.688172  *)

Integer powers are treated differently than floating-point powers. Total tends to be a better way to approach small finite sums.

Answer 3

I want to mention that this sum has a closed form in terms of Mathematica functions

(-1)^n*LerchPhi[-1, 1, 1 + n] - Log[2]

As already carefully pointed out by Michael, the (-1)^n part introduces the small imaginary parts when you apply it to real values. If you are interested in a approximate result, the easiest way is to calculate the above term for integers and then taking N.

f[n_Integer] := N[(-1)^n*LerchPhi[-1, 1, 1 + n] - Log[2]]

f[100]
(* -0.688172 *)

However, calculating LerchPhi should be faster when calculated with approximate numbers in the first place. So the other solution is to put the N inside LerchPhi. Therefore, the better alternative if you want to calculate this for large n might be this

f2[n_Integer] := (-1)^n*LerchPhi[-1, 1, 1 + N[n]] - Log[2]

A quick comparison shows, that I can easily calculate 2^20 in half a second

f2[2^20] // AbsoluteTiming
(* {0.508786, -0.693147} *)

while the exact computation of f needs about 300x the time of this

f[2^20] // AbsoluteTiming
(* {147.191, -0.693147} *)

Runtime-wise, nothing will beat NSum if it implicitly contains a more expensive function. Therefore, my last point is to mention that you can easily split your sum into to parts: All positive terms and all negative terms. With this, you get completely rid of the (-1)^n term and no Chop is necessary

nn = 100;
NSum[1/n, {n, 2, nn, 2}] - NSum[1/n, {n, 1, nn - 1, 2}]
(* -0.688172 *)