Question:
Find the sum $$I=\sum_{n=1}^{\infty}\dfrac{L_{n}(2)}{n^4}=?$$ where $$L_{n}(k)=1-\dfrac{1}{2^k}+\dfrac{1}{3^k}-\cdots+\dfrac{(-1)^{n-1}}{n^k}$$
since $$L_{n}(2)=1-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots+\dfrac{(-1)^{n-1}}{n^2}$$ so $$I=\sum_{n=1}^{\infty}\dfrac{1-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots+\dfrac{(-1)^{n-1}}{n^2}}{n^4}$$
Then I can't Continue