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Is $\lim_{x \rightarrow 1} \sum_{n=1}^\infty \frac{nx^2}{n^3+x}=\sum_{n=1}^\infty \frac{n}{n^3 +1}$ correct?

I do not know how to solve this question. I wanted to calculate the convergence radius and I thought that if $1$ is less or equal to this, than the statement is correct. But I do not know how to calculate this without being able to separate the $x$ from the terms $c_n$.

And they also ask if this is always true, but then I would answer only if $x$ is smaller than the convergence radius. However, I do not know how to prove this or imply this via actual definitions.

Gary
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notaMathqueen
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  • Jose Carlos Santos's answer is spot on. FWIW, I want to point out that you just wish to know as to when you can interchange the limit and the infinite summation. The following post might help. https://math.stackexchange.com/questions/385470/conditions-for-taking-a-limit-into-an-infinite-sum – TryingHardToBecomeAGoodPrSlvr Jun 10 '21 at 06:57
  • Similar: https://math.stackexchange.com/q/3978369, https://math.stackexchange.com/q/2396348. – Martin R Jun 10 '21 at 07:08

2 Answers2

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Note that the siries $\sum_{n=1}^\infty\frac{nx^2}{n^3+x}$ converges uniformly, by the Weierstrass $M$-test: if $x\in[0,1]$,$$\left|\frac{nx^2}{n^3+x}\right|\leqslant\frac n{n^3}=\frac1{n^2}$$and the series $\sum_{n=1}^\infty\frac1{n^2}$ converges. So,$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\displaystyle\sum_{n=1}^\infty\frac{nx^2}{n^3+x}\end{array}$$is a continuous function. In particular, $f(1)=\lim_{x\to1}f(x)$.

José Carlos Santos
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  1. The series $\sum_{n=1}^\infty \frac{nx^2}{n^3+x}$ is not a power series ! Hence "convergence radius" makes no sense.

  2. Show that for $x \in [0,1]$ we have

$$0 \le \frac{nx^2}{n^3+x} \le \frac{1}{n^2}$$

for all $n$. Weierstraß now shows that the series $\sum_{n=1}^\infty \frac{nx^2}{n^3+x}$ is uniformly convergent on $[0,1].$

Can you proceed ?

Fred
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