Investigate the convergence of $\sum_{n=1}^\infty$ $n+1\over (n^3+2)^{1\over 2}$. Investigate the convergence of $\sum_{n=1}^\infty$ $n+1\over (n^3+2)^{1\over 2}$.
Since $n+1<n^2$ and $(n^3+2)^{1\over 2}>n^{3\over 2}$ for all $n\ge2$, then we can compare the above seres to $n^2\over n^{3/2}$$=$$n^{1\over 2}$ which is a divergent series. Thus by the comparison test we know that the original series is divergent.
Does this make sense. Any hints/help is greatly appreciated.
 A: It is true that the series is divergent, but  $$n^{1/2}>  \dfrac{n+1}{(n^3 + 2)^{1/2}}$$ 
so we'll find a divergent series $\sum b_n$ such that $$\dfrac {n+1}{(n^3 + 2)^{1/2}} > b_n$$ Note that $\dfrac{n+1}{(n^3 + 2)^{1/2}}\,>\,\dfrac 1{n^{1/2}}$, and since  $\sum \dfrac 1{n^{1/2}}$ diverges by the $p$-test, so too does our series.
A: Intuitively, the series diverges, since $\frac{n+1}{(n^3+2)^{1/2}} \approx \frac{n}{n^{3/2}} = \frac{1}{\sqrt{n}}$. So, we need to bound the series below by something that also diverges; a clever direct comparison is
$$ \frac{n+1}{(n^3+2)^{1/2}} > \frac{n+1}{(n(n+1)^2)^{1/2}} = \frac{n+1}{(n+1)\sqrt{n}} = \frac{1}{\sqrt{n}}. $$
And of course, the divergence of $\sum \frac{1}{\sqrt{n}}$ is established by the "p-test".
A: Your way isn't correct cause you can't raise the numerator since then each term is bigger then the one from original series,also you can't make the denominator smaller since then that number is bigger then the original one.So you either make numerator smaller and/or denominator bigger.
Anyway notice $n^3+2\leq n^4$ for $n\geq2$ so you have that
$$\frac{n+1}{(n^4)^{1/2}}=\frac{1}{n}+\frac{1}{n^2}\geq\frac{n+1}{(n^3+2)^{1/2}}$$
A: Let us consider the consider a plot of the functions corresponding to the sequence $(a_n) = \displaystyle \bigg(\frac{1}{n^\frac{1}{2}}\bigg)$ as well as our given sequence $(b_n)=\bigg( \frac{n+1}{(n^3 +2)^\frac{1}{2}}\bigg)$ respectively.

Looking at the plot given above we can see that, eventually $$\frac{1}{n^\frac{1}{2}} < \frac{n+1}{(n^3 +2)^\frac{1}{2}}$$
Notice, however, $\displaystyle \sum_{n=1}^\infty\frac{1}{n^\frac{1}{2}}$ diverges (read up $p$-series).
We can thus, using the Limit Comparison Test, conclude that $\displaystyle \sum_{n=1}^\infty \frac{n+1}{(n^3 +2)^\frac{1}{2}}$ diverges.
