How do you test the series #Sigma rootn(n)/n^2# from n is #[1,oo)# for convergence?

Question

Scarlett Allison · Accepted Answer

undefined You can test the series $ \sum \frac{\sqrt{n}}{n^2} $ for convergence using the Limit Comparison Test or the Ratio Test. Let's use the Ratio Test:

1. Take the ratio of consecutive terms:
   $ \frac{a_{n+1}}{a_n} = \frac{\frac{\sqrt{n+1}}{(n+1)^2}}{\frac{\sqrt{n}}{n^2}} $
  
2. Simplify the ratio:
   $ \frac{a_{n+1}}{a_n} = \frac{\sqrt{n+1}}{(n+1)^2} \times \frac{n^2}{\sqrt{n}} $

3. Simplify further:
   $ \frac{a_{n+1}}{a_n} = \frac{n^2}{n^2+2n+1} \times \frac{n^2}{\sqrt{n}} $

4. Take the limit as $ n $ approaches infinity:
   $ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{n^4}{(n^2+2n+1)\sqrt{n}} $

5. Simplify and evaluate the limit:
   $ \lim_{n \to \infty} \frac{n^4}{(n^2+2n+1)\sqrt{n}} = \lim_{n \to \infty} \frac{n^3}{\sqrt{n}(1+2/n+1/n^2)} $
  
6. Since the denominator grows faster than the numerator, the limit is 0.

7. By the Ratio Test, if the limit is less than 1, the series converges.

Therefore, $ \sum \frac{\sqrt{n}}{n^2} $ converges.

Christopher Agresta · Answer

We can conclude that:

#sum_(n=0)^oo root(n)(n)/n^2#

is convergent by direct comparison with #sum_(n=0)^oo 1/n^2#.

Note that: #root(n)(n)/n^2 = n^(1/n)/n^2 = (e^lnn)^(1/n)/n^2 = (e^(lnn/n))/n^2# Now as: # ln n < n => lnn/n < 1 => e^(lnn/n) < e # we have: #root(n)(n)/n^2 < e/n^2# Based on the p-series test we know that: #sum_(n=0)^oo e/n^2 = e*sum_(n=0)^oo 1/n^2# is convergent, so that by direct comparison we can conclude that our series is convergent.