hello, i need help with one more problem:
Use the Integral Test to show that
∞
Σ [(ln n)^k]/n^2 converges for all exponents k. You may use that
n=2
∞
integral (u^k)(e^-u) du converges for all k
1
thanks!
So the integral test says that if the integral of the series converges, then the series converges as well. To integrate this, start with the substitution u=ln(n) -> $\displaystyle du=\frac{1}{n}dn$. So the numerator is simply u^2 and from the du term we have (1/n)dn There is still an extra n unaccounted for though in the denominator. So notice that e^u=n by our original substitution. Thus we can write that this integral is also $\displaystyle \frac{u^k}{e^u}du$ and it seems you're allowed to simply state that this converges.