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

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

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 $(2)$.