I tried the alternating series test but it did not work because the limit does not exist.

Ratio test gave me 1 so that was no help either.

  • 1
    $\begingroup$ $\displaystyle\lim_{n\to\infty}\frac1{\sqrt{n+1}}=0$ $\endgroup$ – robjohn Apr 15 '13 at 16:34

This is an alternating series. To verify that, we need to check three items.

  1. Signs alternate.

  2. The terms go down in absolute value. In symbols, $|a_{n+1}|\le |a_n|$ for all $n$.

  3. The terms have limit $0$. In symbols, $\lim_{n\to\infty}a_n=0$.

It is easy to see that in our case, all three conditions are met.

Remarks: In the post, it is stated that the limit does not exist. You probably meant that $\sum \frac{1}{\sqrt{n+1}}$ does not exist. That is true, but it has nothing to do with the alternating series test. It does imply that your series does not converge absolutely, it only converges conditionally. Informally, that means that if it weren't for the minus signs, we would not have convergence.

Note that we can use the alternating series test if the conditions we listed are true after a while. For example, you could remove the first $100$ minus signs and still have convergence. Or up to the $100$-th term the absolute values don't do down, but after that they do.

  • $\begingroup$ So you're saying the alternating series test should've worked? $\endgroup$ – user72708 Apr 15 '13 at 16:33
  • $\begingroup$ @user72708 the signs of the series is + , -, +, - .... $\endgroup$ – Lost1 Apr 15 '13 at 16:34
  • $\begingroup$ Perfectly, and immediately. In principle there are three conditions to check. They are all easy. $\endgroup$ – André Nicolas Apr 15 '13 at 16:34
  • $\begingroup$ Oh. I see it now. Thank you. My nervousness is killing me. $\endgroup$ – user72708 Apr 15 '13 at 16:36

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