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I am curious about this sum because (as wolfram alpha tells me) it simplifies to a rational number:

$$\sum_{n= 1}^{\infty} \frac{(-1)^n\ln(n)}{n(n+1)} = 0.063254$$

I found this interesting because I did not expect this complicated sum to converge so nicely.

My question is two fold:

1) Is there a general technique that is used to solve sums that involve $\ln(n)$?

2) What hints can you give me to help solve this sum?

user3760593
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    Are you sure it's not just first few digits of a (likely irrational) numerical value? – Yuriy S Apr 07 '18 at 22:13
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    Pari/GP gives sumalt(n=1, (-1)^n*log(n)/n/(n+1)) = 0.063253969010204084743376 and I trust Pari more than Wolfram for this type of sums, see e.g. my question https://math.stackexchange.com/questions/476492/value-of-sum-n-0-infty-frac-1n-lnn2 – gammatester Apr 07 '18 at 22:23
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    Here's 20 digits WA gives for the integral representation I found 0.063253969010204080235. A little different from gammatester's result, but still doesn't seem to be the rational number the OP assumed. Here's the link: http://www.wolframalpha.com/input/?i=NIntegrate%5BHypergeometricPFQ%5B%7B2,2,1%2B1%2Ft%7D,%7B4,2%2B1%2Ft%7D,-1%5D%2F(1%2Bt)%2F6,%7Bt,0,1%7D,WorkingPrecision-%3E20%5D – Yuriy S Apr 07 '18 at 22:38
  • I see, the sum does not converge to a rational number. I should have tried writing a program to check wolfram's answer. – user3760593 Apr 07 '18 at 22:43
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    Here's the value WA gives for another integral representation. Apparently its numerical algorithms are a little faulty: 0.063253969010204084773. http://www.wolframalpha.com/input/?i=NIntegrate%5B(Gamma%5B0,-Log%5Bx%5D%5D)((2%2Bx)Log%5B1%2Bx%5D-2x)%2Fx%5E3,%7Bx,0,1%7D,WorkingPrecision-%3E20%5D – Yuriy S Apr 07 '18 at 22:54
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    Don't say Wolfram is faulty. Wolfram just computed this to 6 places. Probably you can tell Wolfram to tell you the result to more digits. – GEdgar Apr 08 '18 at 00:27
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    @GEdgar, I explicitly asked it for 20 digits, I guess WorkingPrecision doesn't always lead to the same number of correct digits in the answer – Yuriy S Apr 08 '18 at 00:43
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    Partial fractions and some stuff gives me this: \begin{align}\sum_{n=1}^\infty\frac{(-1)^n\ln(n)}{n(n+1)}&=\sum_{n=1}^\infty\frac{(-1)^n\ln(n)}n-\sum_{n=1}^\infty\frac{(-1)^n\ln(n)}{n+1}\&=\frac d{dx}\sum_{n=1}^\infty\frac{(-1)^nn^x}n\bigg|{x=0}-\sum{n=1}^\infty\frac{(-1)^n\ln(n)}{n+1}\&=\eta'(1)-\sum_{n=1}^\infty\frac{(-1)^n\ln(n)}{n+1}\&=\gamma\ln(2)-\frac12[\ln(2)]^2-\sum_{n=1}^\infty\frac{(-1)^n\ln(n)}{n+1}\end{align} – Simply Beautiful Art Apr 08 '18 at 01:02

1 Answers1

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We can find integral representations of this series. The most direct way would be to represent the logarithm as an integral:

$$\ln n=(n-1) \int_0^1 \frac{dt}{1+(n-1)t}$$

Interchanging integration and summation, we can write the series as:

$$\sum_{n= 2}^{\infty} \frac{(-1)^n\ln n}{n(n+1)}=\int_0^1 dt \sum_{n= 2}^{\infty} \frac{(-1)^n (n-1)}{n(n+1)(t n-t+1)}$$

The inner sum can be found in hypergeometric form the following way. First we shift the index:

$$\sum_{n= 2}^{\infty} \frac{(-1)^n (n-1)}{n(n+1)(t n-t+1)}=\sum_{n= 0}^{\infty} \frac{(-1)^n (n+1)}{(n+2)(n+3)(t n+t+1)}$$

Now we find the $0$th term and the ratio of successive terms:

$$c_0=\frac{1}{6(1+t)}$$

$$\frac{c_{n+1}}{c_n}=\frac{(n+2)(n+2)\left(n+\frac{1}{t}+1 \right)}{(n+4)\left(n+\frac{1}{t}+2 \right)} \frac{(-1)}{n+1}$$

Which makes the series equal to:

$$\sum_{n= 2}^{\infty} \frac{(-1)^n (n-1)}{n(n+1)(t n-t+1)}=\frac{1}{6(1+t)}~ {_3 F_2} \left(2,2,\frac{1}{t}+1;4,\frac{1}{t}+2;-1 \right)$$

This gives us an integral representation:

$$\sum_{n= 2}^{\infty} \frac{(-1)^n\ln n}{n(n+1)}= \frac{1}{6} \int_0^1 {_3 F_2} \left(2,2,\frac{1}{t}+1;4,\frac{1}{t}+2;-1 \right)\frac{dt}{1+t} \tag{1}$$

We can use Euler's integral transform to reduce the order of the hypergeometric function and obtain a double integral in terms of Gauss hypergeometric function ${_2 F_1} (2,2;4;-x)$, which in this case has elementary form. Then we integrate w.r.t. $t$ and obtain another integral representation:

$$\sum_{n= 2}^{\infty} \frac{(-1)^n\ln n}{n(n+1)}=\int_0^1 \Gamma(0,-\ln x) \left((2+x) \ln (1+x)-2x \right) \frac{dx}{x^3} \tag{2}$$

Where the incomplete Gamma function appears.

Using Simply Beautiful Art's result, we can also write:

$$\sum_{n= 2}^{\infty} \frac{(-1)^n\ln n}{n(n+1)}=\gamma \ln 2-\frac{\ln^2 2}{2} -\frac{1}{3} \int_0^1 {_3 F_2} \left(2,3,\frac{1}{t}+1;4,\frac{1}{t}+2;-1 \right)\frac{dt}{1+t} \tag{3}$$

The integral in $x$ will be a little more complicated than $(2)$.

Yuriy S
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