SumConvergence[((-1)^n)/(Sqrt[n] + (-1)^n), n] returns True in Version 10.2?

Question

Bug introduced in 10.4.1 or earlier and fixed in 11.0.0

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I claim that the series $\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt{n}+(-1)^n}$ diverges. To see this, rewrite the $n^{th}$ term as follows: \begin{equation*} \frac{(-1)^n}{\sqrt{n}+(-1)^n} = \frac{(-1)^n}{\sqrt{n}+(-1)^n} \times \frac{\sqrt{n}+(-1)^{n+1}}{\sqrt{n}+(-1)^{n+1}} \end{equation*} \begin{equation*} = \frac{(-1)^n\sqrt{n}+(-1)^{2n+1}}{n+(-1)^{2n+1}} = \frac{(-1)^n\sqrt{n}}{n-1} + \frac{-1}{n-1} \end{equation*} Thus, \begin{equation*} \sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt{n}+(-1)^n} = \sum_{n=2}^{\infty}\frac{(-1)^n\sqrt{n}}{n-1} - \sum_{n=1}^{\infty}\frac{1}{n} \end{equation*} Now, we have the difference of two series. The first series, by the Alternating Series Test, is convergent, and the second is the well known divergent harmonic series.

However,

    SumConvergence[((-1)^n)/(Sqrt[n] + (-1)^n), n]

returns True, and

    N[Sum[((-1)^n)/(Sqrt[n] + (-1)^n), {n, 2, Infinity}]]

returns -2.70244.

To add some evidence for the divergence of the series, the code

    pow10 = 10;
    expo = 1;
    While[ expo <= 6,
      Print [N[Sum[((-1)^n)/(Sqrt[n] + (-1)^n), {n, 2, pow10}]]];
      expo++;
      pow10 *= 10;
    ]

returns

    -1.74377
    -4.21214
    -6.55378
    -8.86764
    -11.1737
    -13.4774

I see in another post that there was an issue with SumConvergence in version 10.0.0.0, but I am using Version 10.2. Any insights here?

Answer 1

Another way to rewrite the series:

Sum[((-1)^n)/(Sqrt[n] + (-1)^n), {n, 2, 1001}] ==
 Sum[1/(Sqrt[2 n] + 1) - 1/(Sqrt[2 n + 1] - 1), {n, 1, 500}]

True

SumConvergence[((-1)^n)/(Sqrt[n] + (-1)^n), n]

True

SumConvergence[1/(Sqrt[2 n] + 1) - 1/(Sqrt[2 n + 1] - 1), n]

False

( v 10.1 )

This is not rearranging terms, but grouping in pairs; the series are the same:

Answer 2

Wow. Just wow. So slightly different $\sum_{n=2}^{\infty} \dfrac{(-1)^n}{n+(-1)^n}$ is a classic example of (conditionally) convergent series for which Leibniz test fails. Leibniz test's monotonicity is NOT a necessary condition, thus the test itself is only sufficient, but not necessary! This is a very important result, because the test includes main nessessary condition for all series!

As for your variant of this series... It is also an example. See (diverges by Tailor): https://math.stackexchange.com/a/4221590 It is actually a bad idea to use something like this (but it is correct, sum of divergent and convergent series is always divergent) since if it is conditionally convergent you can always rearrange the terms to make it divergent (Riemann theorem) and thus sometimes vice versa is possible. So... Use Tailor instead and it is divergent, yes: https://math.stackexchange.com/a/2423137

There are other examples of such series with now absolute convergence. https://math.stackexchange.com/a/1073294/756502

I am not surprised Mathematica fails with these. They should really be remembered. Even this fails, but on Alpha it works:

a[n_] := ((-1)^n)/(n + (-1)^n);
SumConvergence[a[n], n]

Also see: https://math.stackexchange.com/a/1858185/756502