# compute the following limit of this series (derivate in $0$)

Let's consider the function $f(x)=\sum_{n=1}^{\infty}\frac{1}{n^{2}+n^{3}x^{2}}$ , I proved that $f'(x)=\sum_{n=1}^{\infty}(\frac{1}{n^{2}+n^{3}x^{2}})'=\sum_{n=1}^{\infty}-\frac{2x}{n(nx^{2}+1)^{2}}$ on $(0,\infty)$ I want to prove that $f'(0)=0$ to prove that this equality holds on all $\Bbb R$. i.e to prove the following: $$\left| {f'\left( 0 \right)} \right| = \left| {\frac{1} {h}\left( {f\left( h \right) - f\left( 0 \right)} \right)} \right| = \left| {\frac{1} {h}\sum\limits_{k = 1}^\infty {\frac{1} {{k^2 + k^3 h^3 }} - \frac{1} {{k^2 }}} } \right| = \left| {\frac{1} {h}\sum\limits_{k = 1}^\infty {\frac{1} {{k^2 }}\frac{1} {{1 + h^3 k}} - 1} } \right| = \left| {\sum\limits_{k = 1}^\infty {\frac{{h^2 }} {{k\left( {1 + h^3 k} \right)}}} } \right|$$ I have to prove that $$\left| {\sum\limits_{k = 1}^\infty {\frac{{h^2 }} {{k\left( {1 + h^3 k} \right)}}} } \right| \to 0$$ But I don't know how

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The $h^3$ in the denominator should be corrected to $h^2$. – Sangchul Lee Nov 23 '12 at 5:52

Clearly no ambiguity arises if we assume $h > 0$. Then by AM-GM inequality,

$$1 + h^2k \geq 2\sqrt{h^2 k} = 2h \sqrt{k}.$$

Thus we have

$$\sum_{k=1}^{\infty} \frac{h^2}{k(1 + h^2 k)} \leq \sum_{k=1}^{\infty} \frac{h^2}{k \cdot 2 h \sqrt{k}} = \frac{h}{2} \sum_{k=1}^{\infty} \frac{1}{k^{3/2}},$$

which clearly converges to $0$ as $h \to 0$.

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@Chris'ssister, Can you come in to the chatroom? – Sangchul Lee Nov 23 '12 at 8:52
I'm in the chatroom. – user 1618033 Nov 23 '12 at 9:14