Recent content by maxitis

so the series converges right? irrelevant, the: \sum_{n=1}^\infty\frac{(-1)^n}{n}=\sum_{n=1}^\infty\frac{1}{2*n}-\frac{1}{2*n-1} isn't that correct?

The absolute value does not converge. \sum_{n=2}^\infty\frac{1}{ln(n)}>=\sum_{n=2}^{\infty}\frac{1}{x-1} So i think we can't use that (i write right the infinity but it always put a space between)

Sequence Convergence \sum_{n=2}^\infty\frac{(-1)^n}{ln(n)} I have tried some comparisons bot not conclude: \sum_{n=2}^\infty\frac{(-1)^n}{ln(n)}<=\sum_{n=2}^\infty\frac{(-1)^n}{1} \sum_{n=2}^\infty\frac{(-1)^n}{x-1}<=\sum_{n=2}^\infty\frac{(-1)^n}{ln(n)} Somebody having any insights...

The numerator should be zero. So we have: f(2)-5=0 Am i right? Something about before, Why you said that there are infinite number of functions f that can satisfy that limit? One is f(x)=5(x-1), can you give me another example? Thank you

And why is that a problem? Both parts will be equal to 0.

I have the limit: \lim_{x\rightarrow +2} {\frac{f(x)-5}{x-2}}=5 And i want to find the: \lim_{n\rightarrow +2} {f(x)} Can i say that f(x)-5=5*(x-2) And then find the limit? Thank you