∫dx/(x^2*lnx)
What I`ve seen on the web but I don`t think is right:
u= lnx *** what we have here isn't lnx but (lnx)^-1... This is why I doubt that's the right solution
du = dx/x
dv = dx/x^2
v = -1/x
=-lnx/x + ∫dx/x^2
=-lnx/x - 1/x + C
Let me know if it is correct, thanks!
Quoting my notes:
If ∫[1,∞[ f(x) is convergent <=> Ʃ∞n=1 an is convergent where an = f(x)
So the goal is to prove the integral is convergent to find out if the serie is converging.
Just making sure I understand you correctly (I am natively speaking french).
The serie Ʃ∞n=1 ne^(-2n) = e^-2 + 2e^(-4) + ... + ne^(-2n)
I can clearly see here the serie is decreasing and convergent because it gets closer to a number since an > an+1.
I would like to prove it with the...
Been a long time I had my integral class so I forgot almost everything I knew... I need to integrate to see if the serie converge (limn→∞ an = 0). Thus, there is a theorem of the integral, if you evaluate the limit of the integral of a serie when it tends to the infinite minus when x=1 you can...