alternating series of (-1)^n (ln(n))/n^2 can i compare the series to a 1/n^2 convergent series and then say that by limit comparison test the series absolutely converges?

Question

nottim · Answer

What subject is this?

anonymous · Answer

math...

nottim · Answer

wow. then I must be super dumb.

anonymous · Answer

? amm is calculus 2 ... alternating series.. ... .. amm i think im still doing math lol

zarkon · Answer

$$\ln(n)\le \sqrt{n}$$

zarkon · Answer

$$\frac{\ln(n)}{n^2}\le\frac{\sqrt{n}}{n^2}=\frac{1}{n^{3/2}}$$

experimentx · Answer

Yup, Zarkon is right, 1/n^(3/2) converges
so left hand side must also converge.
we can show that (-1)^n * (ln(n))/n^2 converges if (ln(n))/n^2 converges, since absolute value is greater than the other alternating value

anonymous · Answer

ok but why compareto to ((n)^1/2))/(n^2)???

zarkon · Answer

why not...it works

experimentx · Answer

because ((n)^1/2))/(n^2) is a convergent series and your series is less than it

anonymous · Answer

got it thanks guys...