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

Question

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?

Answer 1

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.

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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 $(2)$.