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$$\sum_{n=0}^\infty \frac{\sqrt n}{n+7}$$

I tried to use the comparison test and I know I should use a lower comparison since it's very much likely diverges but had problems lowering the square root. (According the estimation I did)

Could you help me with the comparison test? (or any other hint if you would use other tests)

EDIT: The exam actually asks if it diverges or convergent but I know it's diverges, just can't prove it yet.

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    $\begingroup$ In this, as in most cases, the best form of the comparison test to use is the limit comparison test. $\endgroup$ – Angina Seng May 4 '17 at 6:23
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    $\begingroup$ Hint: $\;\sqrt{n} \ge 1\,$ for $\,n \ge 1\,$, then remember that the harmonic series diverges. $\endgroup$ – dxiv May 4 '17 at 6:24
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    $\begingroup$ the limit comparison test to $\sum \frac {1}{\sqrt {n}}$ would be the place that I would start....or the direct comparison test to $\sum \frac {1}{2\sqrt {n}}$ $\endgroup$ – Doug M May 4 '17 at 6:24
  • $\begingroup$ Hint: for large $n$, the constant $7$ becomes negligible and you end-up with the simplified term $1/\sqrt n$, which decays slower than the harmonic series. $\endgroup$ – Yves Daoust May 4 '17 at 6:26
  • $\begingroup$ Thanks a lot guys, I overlooked the fact that it's enough to write $$\sqrt n / n $$I tried to lower the square root as well but I was wrong. $\endgroup$ – 2b1c May 4 '17 at 6:31
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Ignoring the first term and shifting the index,

$$\sum_{n=1}^\infty\frac{\sqrt n}{n+7}=\sum_{n=8}^\infty\frac{\sqrt{n-7}}{n}>\sum_{n=8}^\infty\frac1{n}.$$

Hence the series diverges.

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