Analyze convergence of the series $\sum\limits_{n=1}^\infty\frac{1}{n(n^\frac14-n^\frac13+n^\frac12)}$ $$\sum_{n=1}^\infty\frac{1}{n(n^\frac{1}{4}-n^\frac{1}{3}+n^\frac{1}{2})}$$
It is possible to use the ratio test or the root test or any of them without give opposite answers ? There is a case in which a method give an answer and another other answer? Which of them is a better strategy for this one? Does this one converge or diverge? How I recognize which method I ought use?
Sorry for asking so many questions.
 A: Note that for large $n$ we have 
$$
\frac{1}{n(n^{\frac{1}{4}}-n^{\frac{1}{3}}+n^{\frac{1}{2}})}\sim \frac{1}{n^{3/2}}
$$
and the above sequence summed converges
A: Take a look at $(n^\frac14-n^\frac13+n^\frac12)$. As $n \to \infty$, $n^\frac12$ will get larger exponentially faster than $n^\frac14$ and $n^\frac13$ since it has a larger exponent, $\frac12$. As a result, $(n^\frac14-n^\frac13+n^\frac12)$ starts to look like just $n^\frac12$ as the series moves to its larger terms; $n^\frac14$ and $n^\frac13$ become inconsequential.
This means that as
$$
\sum_{n=1}^\infty \frac1{n(n^\frac14-n^\frac13+n^\frac12)}
$$
approaches its infinitieth term, it will begin to behave the same way as
$$
\sum_{n=1}^\infty \frac1{n(n^\frac12)}
$$
which is the same as
$$
\sum_{n=1}^\infty \frac1{n^{1+\frac12}}
=\sum_{n=1}^\infty \frac1{n^\frac32}
$$
And this series converges by the p-series test which states that for a series
$$
\sum_{n=1}^\infty \frac1{n^p}
$$
the series will converge as long as $p>1$.
Because this series converges, you can confidently assert that the original series converges because its end behavior is the same, and it's the end behavior of a series, how it behaves as it reaches its infinitieth term, that determines whether or not it converges.
A: Just for the fun.
Consider $$A=\frac{1}{x(x^\frac{1}{4}-x^\frac{1}{3}+x^\frac{1}{2})}$$ and make $x=y^{12}$ to get $$A=\frac{1}{y^{15} \left(y^3-y+1\right)}=\frac{1}{y^{18} \left(1-\frac 1{y^2}+\frac 1 {y^3}\right)}$$ Now, make the long division (or use Taylor series) to get $$\frac{1}{ 1-\frac 1{y^2}+\frac 1 {y^3}}=1+\frac 1{y^2}-\frac 1 {y^3}+\frac 1 {y^3}+\frac 1 {y^4}+O\left(\frac 1 {y^5} \right)$$ making $$A=\frac 1 {y^{18}}+\frac 1 {y^{20}}-\frac 1 {y^{21}}+\frac 1 {y^{22}}+O\left(\frac 1 {y^{23}} \right)$$ Back to $x$ $$A=\frac 1 {x^{3/2}}+\frac 1 {x^{5/3}}-\frac 1 {x^{7/4}}+\frac 1 {x^{11/6}}+\cdots$$ Replace $x$ by $n$ and prepare the summations : you face convergent series.
A: Hint: observe that
$$
\frac{1}{n(n^\frac{1}{4}-n^\frac{1}{3}+n^\frac{1}{2})}
= \frac{1}{n \, n^\frac{1}{2}(1  -n^{-1/6} +n^{-1/4})}
$$
and that the term in brackets goes to $1$ as $n\to +\infty$.
