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$$\sum_1^\infty \frac{nx^n}{n^3+x^{2n}}$$

Here is what I've done :

$$|\frac{nx^n}{n^3+x^{2n}}|<|\frac{nx^n}{n^3}|=|\frac{x^n}{n^2}|$$$

Since series on $RHS$ would converge for $|x|<1$, then whenever $|x|<1$ is satisfied $LHS$ would converge. However answer is

LHS converges $\forall \epsilon R$ real numbers.

1 Answers1

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Case 1. Assume $|x|\le1$, then $$ \sum_1^\infty \left|\frac{nx^n}{n^3+x^{2n}} \right|\le \sum_1^\infty \frac{n|x|^n}{n^3} \le \sum_1^\infty\frac{1}{n^2} <\infty $$ and the given series is absolutely convergent.

Case 2. Assume $|x|>1$, then $$ \sum_1^\infty \left|\frac{nx^n}{n^3+x^{2n}} \right|\le \sum_1^\infty \frac{n|x|^n}{|x|^{2n}} = \sum_1^\infty \frac{n}{|x|^{n}} <\infty $$ and the given series is absolutely convergent, where we have used this standard result.

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Olivier Oloa
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  • For $|x|>1$ a very elementary finish can be: When $n\geq 3$ and $|x|=1+y $ with $ y>0$ we have: $|x|^n=(1+y)^n\geq 1+y\binom {n}{1}+y^2\binom {n}{2}+y^3\binom {n}{3}>y^3 \binom {n}{3},$ so $0<n/|x|^n<ny^{-3} /\binom {n}{3}=6y^{-3}/((n-1)(n-2))$...............+1 – DanielWainfleet Mar 07 '17 at 20:54