# How can I calculate the sum of this series correctly?

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]

• NSum[((-3)^n/(3^n + (-2)^n) 1/n), {n, 1, Infinity}] does the job, outputting -3.09333. Aug 22, 2020 at 5:24
• @user64494 Thank you very much, but I also want to know its exact value. Aug 22, 2020 at 5:25
• I have strong doubts concerning a closed-form expression for the value of the sum under consideration. Aug 22, 2020 at 5:29
• I don't know such proofs for either infinite series and improper integrals. Aug 22, 2020 at 6:28
• @Montevideo Proving that the limit of a sum does not have a closed-form expression is not something you can expect out of Mathematica. Aug 22, 2020 at 9:28

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

• Amazing, very nice solution! So the sum can be computed in closed form. Aug 22, 2020 at 21:56
• 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. Aug 22, 2020 at 22:09
• @yarchik This number Log[QPochhammer[q, q^2]/QPochhammer[-1, q^2]] /. q -> 2/3 should belong to irrational number. Aug 22, 2020 at 22:22
• $$\log \left(\frac{\left(\frac{2}{3};\frac{4}{9}\right)_{\infty }}{\left(-1;\frac{4}{9}\right)_{\infty }}\right)$$ is not a closed-form expression up to en.wikipedia.org/wiki/Closed-form_expression . Aug 23, 2020 at 12:33
• In fact, the sum of the series is expressed in terms of infinite products (see reference.wolfram.com/language/ref/QPochhammer.html). Aug 23, 2020 at 12:58

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.

• 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. Aug 23, 2020 at 15:59
• 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? Aug 24, 2020 at 5:39