Does 1 sqrt converge

Now, i seek your solemn opinions, gentlemen.( original post by munn) you could show whether or not converges to 1.For x > e2 we have that lnx > 2 and then df dx < 0, so that f (x) is monotone decreasing.We can see that the series is divergent by analyzing the partial sums:The asymptotic mean and variance of y n is unity, and from the clt we have:

1 n + n + 1 ≥ 1 4 n ≥ 0.Ask question asked 8 years ago.Hence by the integral test sum 1/sqrt(n) diverges.+ (√n −√n − 1) + (√n +1 − √n) = √n + 1.Like walking up to me and telling me the answer to 13 cubed.

I think that i should use something to do with 1/sqrt (x^6).Lim int 1/x dx = lim log x = infinity.Viewed 134 times 3 $\begingroup$ does the sequence $\sqrt[n]{1+a^n}$, where $0

The tob and bottom don't even converge to a limit.Since the other posted proofs all seem to be based on the integral test, i'll give a solution by using a comparison test.Integral of 1/sqrt (x^6+1) from 0 to infinity.