Using the integral test, how do you show whether $(\frac{1}{\sqrt{n + 1}})$ $(1/sqrt (n+1))$ diverges or converges?

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Using the integral test, how do you show whether $(\frac{1}{\sqrt{n + 1}})$ $(1/sqrt (n+1))$ diverges or converges?

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Answer 1 · 2015-06-11T00:57:47

I would assume that this goes from $n = 1$ $n = 1$ to $\infty$ $oo$ ? The domain must be $[k, \infty]$ $[k,oo]$ in order for this to work. If so, I've already answered this here:
http://socratic.org/questions/using-the-integral-test-how-do-you-show-whether-sum-1-sqrt-n-1-diverges-or-conve#151880

Overall, it diverges, because the integral evaluates as $\infty$ $oo$ . The gist of it is, integrate it replacing $n$ $n$ with $x$ $x$ , and show that it will result in $\infty$ $oo$ when evaluated from $1$ $1$ to $\infty$ $oo$ .

You should also know that this is a "p-series":

$\infty \sum n = 1 \frac{1}{{(n \pm k)}^{p}}$ $sum_(n = 1)^(oo) 1/((npmk)^p)$
where $k in RR$ ( $k$ is in the set of real numbers)

If $p > 1$ , the sum will converge, whereas if $p <=1$ , the sum will diverge. When $p = 1$ , it's called the harmonic series

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Using the integral test, how do you show whether $(\frac{1}{\sqrt{n + 1}})$ $(1/sqrt (n+1))$ diverges or converges?

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Using the integral test, how do you show whether $(\frac{1}{\sqrt{n + 1}})$ $(1/sqrt (n+1))$ diverges or converges?