This is an addition on a question I asked earlier. I was given the following functions: $$g(x)=\frac{x}{1+x^{4/3}}\ln(1+\ln x)$$ $$f(x)=g(2x-1)-g(2x)$$
"I" (with help) showed that:
$$2\int_{a}^\infty f(x)dx =\int_{2a-1}^{2a}g(x)dx$$
Now the question is whether it is possible to show that$$S_N=\sum_{k=1}^{2N}(-1)^{k-1}g(k)=\sum_{k=1}^{N}f(k)$$ I understand that we are going from an integral to a partial sum. And I "feel" that the solution is in the $2N$, Any help and hints welcome, Joe
Just split the sum in two by summing separately odd indices and even indices: $$\sum_{k=1}^{2N}(-1)^{k-1}g(k)=\sum_{j=1}^{N}g(2j-1) -\sum_{j=1}^{N}g(2j)= \sum_{j=1}^{N}(g(2j-1)-g(2j))=\sum_{j=1}^{N}f(j).$$