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$$\sum_{n=1}^\infty \frac{(2n+1)^n}{n^{2n}}$$

I am struggling to come up with something. First thought root test but I don't think that will work since the roots aren't the same? I'm guessing maybe the ratio test will work since it usually does but is there a more efficient test to use here?

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What have you tried so far? Why do you think the root test is inefficient? It provides a simple condition which guarantees the convergence of a an infinite sum. Also, it would be nice if instead of copying the sum, you could put the sum in tex. You can easily see how to do it in your previous posts. –  JavaMan Jul 27 '11 at 6:04

4 Answers 4

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The root test should work, since $$ \sqrt[n]{\frac{(2n+1)^n}{n^{2n}}}=\frac{2n+1}{n^2}. $$ It doesn't matter that the exponents of the numerator and denominator are not quite the same.

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Hint: for any $n \geq 6$, $$ \frac{{(2n + 1)^n }}{{n^{2n} }} \le \frac{{(3n)^n }}{{(n^2 )^n }} = \bigg(\frac{3}{n}\bigg)^n \le \bigg(\frac{1}{2}\bigg)^n . $$

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So, comparison test works here, noting that $\sum\limits_{n = 1}^\infty {(\frac{1}{2})^n } < \infty $. –  Shai Covo Jul 27 '11 at 6:30

HINT:

Both the ratio test (with a few generous inequalities) and the root test (definitely the best bet) work here.

Also, there is a general order that I think of and that I told my students when they tested for convergence. It is by no means infallible. Here's what I would say, and in this order:

  • Write out a few terms! Get a feel for the series.
  • Make sure the limit of the terms goes to zero.
  • If it's integrable, use the integral test.
  • If it alternates/telescopes, try the appropriate alternating/telescoping series tests.
  • If it has a factorial, use the ratio test.
  • If it has things raised to nth powers (like this one), use the root test.
  • Use comparison - whichever feels more natural (I think basic is easier to see than limit, but so it goes)

I repeat, this is not infallible. Comparison tests can be scary, and some series are brilliantly bounded using astounding combinations of tests and ingenuity.

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You can rewrite it so that the powers are the same: $n^{2n} = (n^2)^n$. Then use the root test.

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Yeah I saw that at first but for some reason in my head I was thinking it was wrong. Thanks –  Ryan Jul 27 '11 at 6:27

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