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I already know that this numbers series $\sum_{n=1}^{\infty} \frac{1}{3^{n}+(-2)^{n}} \frac{(-3)^{n}}{n}$ is convergent.

SumConvergence[((-3)^n/(3^n + (-2)^n) 1/n), n](*True*)

But the following code can not find its limit value, I want to know how I can correctly find the limit value of this series?

Limit[Sum[(1/(3^n + (-2)^n))*((-3)^n/n), {n, 1, m}], m -> Infinity]
Needs["NumericalCalculus`"]
NLimit[Sum[(1/(3^n + (-2)^n))*((-3)^n/n), {n, 1, m}], m -> Infinity]
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    $\begingroup$ NSum[((-3)^n/(3^n + (-2)^n) 1/n), {n, 1, Infinity}] does the job, outputting -3.09333. $\endgroup$
    – user64494
    Aug 22, 2020 at 5:24
  • $\begingroup$ @user64494 Thank you very much, but I also want to know its exact value. $\endgroup$
    – A little mouse on the pampas
    Aug 22, 2020 at 5:25
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    $\begingroup$ I have strong doubts concerning a closed-form expression for the value of the sum under consideration. $\endgroup$
    – user64494
    Aug 22, 2020 at 5:29
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    $\begingroup$ I don't know such proofs for either infinite series and improper integrals. $\endgroup$
    – user64494
    Aug 22, 2020 at 6:28
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    $\begingroup$ @Montevideo Proving that the limit of a sum does not have a closed-form expression is not something you can expect out of Mathematica. $\endgroup$
    – Sjoerd Smit
    Aug 22, 2020 at 9:28

2 Answers 2

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From

Sum[(-1)^n/((1 + (-q)^n) n), {n, 1, Infinity}]

we get via geometric series to

-Sum[(-1)^l Log[1 + (-q)^l], {l, 0, Infinity}]

and from there to

Log[Product[(1 - q^(2 l + 1))/(1 + q^(2 l)), {l, 0, Infinity}]]

which Mathematica calculates as

Log[QPochhammer[q, q^2]/QPochhammer[-1, q^2]]

Then set q = 2/3.

Andreas

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    $\begingroup$ Amazing, very nice solution! So the sum can be computed in closed form. $\endgroup$
    – yarchik
    Aug 22, 2020 at 21:56
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    $\begingroup$ By NSum[(1/(3^n + (-2)^n))*(-3)^n/n, {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 100, PrecisionGoal -> 95, NSumTerms -> 500] one can verify the analytic solution with 100 digits precision. $\endgroup$
    – yarchik
    Aug 22, 2020 at 22:09
  • $\begingroup$ @yarchik This number Log[QPochhammer[q, q^2]/QPochhammer[-1, q^2]] /. q -> 2/3 should belong to irrational number. $\endgroup$
    – A little mouse on the pampas
    Aug 22, 2020 at 22:22
  • $\begingroup$ $$\log \left(\frac{\left(\frac{2}{3};\frac{4}{9}\right)_{\infty }}{\left(-1;\frac{4}{9}\right)_{\infty }}\right) $$ is not a closed-form expression up to en.wikipedia.org/wiki/Closed-form_expression . $\endgroup$
    – user64494
    Aug 23, 2020 at 12:33
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    $\begingroup$ In fact, the sum of the series is expressed in terms of infinite products (see reference.wolfram.com/language/ref/QPochhammer.html). $\endgroup$
    – user64494
    Aug 23, 2020 at 12:58
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The products may also be expressed in terms of elliptic theta functions as

Log[q/16]/8 + 1/2 Log[EllipticTheta[4, 0, q]/EllipticTheta[2, 0, q]]

Does this count as closed expression?

The easiest way is to use the product representations of the theta functions (https://dlmf.nist.gov/20.5):

EllipticTheta[4, 0, q] = Product[(1 - q^(2 k - 1))^2 (1 - q^(2 k)), {k, 1, Infinity}]

and

EllipticTheta[2, 0, q] = 2 q^(1/4) Product[(1 + q^(2 k))^2 (1 - q^(2 k)), {k, 1, 
Infinity}].

Just enter them into

Log[q/16]/8 + 1/2 Log[EllipticTheta[4, 0, q]/EllipticTheta[2, 0, q]]

and you have it.

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  • $\begingroup$ No, this is not a closed-form expression up to en.wikipedia.org/wiki/Closed-form_expression . In fact, you express the sum of the series through the sum of another series. $\endgroup$
    – user64494
    Aug 23, 2020 at 15:59
  • $\begingroup$ Personally, as someone who deal a lot in special functions, I consider theta functions and $q$-functions as closed forms, so I think you don't really need to pay heed to 64494's objections. Perhaps show how the theta functions come out of the original expression to complete this answer? $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Aug 24, 2020 at 5:39

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