# Prove that the following series is diverges!

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

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.

• In this, as in most cases, the best form of the comparison test to use is the limit comparison test. – Angina Seng May 4 '17 at 6:23
• Hint: $\;\sqrt{n} \ge 1\,$ for $\,n \ge 1\,$, then remember that the harmonic series diverges. – dxiv May 4 '17 at 6:24
• the limit comparison test to $\sum \frac {1}{\sqrt {n}}$ would be the place that I would start....or the direct comparison test to $\sum \frac {1}{2\sqrt {n}}$ – Doug M May 4 '17 at 6:24
• 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. – Yves Daoust May 4 '17 at 6:26
• 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. – 2b1c May 4 '17 at 6:31

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