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I am having difficulty finding radius of convergence for following function:

$$\sum\limits_{1}^{\infty}\cfrac{{(4x-5)}^{2n+1}}{n^{^3/_2}}$$

I tried using ratio test as shown below:

$$ \begin{align} \lim_{n\to\infty}\left| \cfrac{\cfrac{{(4x-5)}^{2(n+1)+1}}{(n+1)^{^3/_2}}}{\cfrac{{(4x-5)}^{2n+1}}{n^{^3/_2}}} \right| &= \lim_{n\to\infty}{\left| \left| (4x-5)^2 \right| \cdot \cfrac{n^{^3/_2}}{(n+1)^{^3/_2}} \right| } \\ &= \lim_{n\to\infty}{\left| \left| (4x-5)^2 \right| \cdot \cfrac{\frac{3}{2}n^{^1/_2}}{\frac{3}{2}(n+1)^{^1/_2}} \right| } \quad (\text{using l'hopital's rule}) \end{align} $$

Now I realised that this is getting me nowhere as further applications of l'hopital's rule will simply not reduce the powers of $ n $ and $ n+1 $ any further. Could someone please advise me on how to solve this question?

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For a limit like $$ \lim_{n\to\infty}\frac{n^{1/2}}{(n+1)^{1/2}} $$ rewrite it as $$ \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{1/2} $$

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  • $\begingroup$ Thanks for your help! I realise how simple this question is now! $\endgroup$ – LanceHAOH Sep 23 '16 at 18:12

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