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Which series test may I apply to find if the following series is convergent or divergent? $$\sum_{n=1}^{\infty}\frac{n}{(n^{1/n}+6)^n}$$

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By comparison we have

$$\sum_{n=1}^{\infty}\frac{n}{(n^{1/n}+6)^{n}}\le \sum_{n=1}^{\infty}\frac{n}{6^n}=\frac{6}{25}$$

where the last equality can be obtained by observing that for

$$\begin{align} f(x)&=\sum_{n=1}^\infty x^n\\\\ =\frac{x}{1-x} \end{align}$$

we have

$$\begin{align} xf'(x)&=\sum_{n=1}^\infty nx^{n}\\\\ &=\frac{x}{(1-x)^2} \tag 1 \end{align}$$

Then, set $x=1/6$ in $(1)$.

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  • $\begingroup$ And how did you get that expression from the final sum? That is not obvious. $\endgroup$ – Rory Daulton Mar 31 '16 at 17:57
  • $\begingroup$ @RoryDaulton There are a couple of ways to go. For example, differentiate the function with series representation $f(x)=\sum_{n=1}^\infty x^n=\frac{x}{1-x}=-1+\frac{1}{1-x}$ for get $f'(x)=\frac{1}{(1-x)^2}$. Then, multiply by $x$ and set $x=1/6$. $\endgroup$ – Mark Viola Mar 31 '16 at 18:00
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The $\;n $ - th root test as this is a positive series

$$\sqrt[n]{\frac n{\left(\sqrt[n]n+6\right)^n}}=\frac{\sqrt[n]n}{\sqrt[n]n+6}\xrightarrow[n\to\infty]{}\frac1{1+6}=\frac17<1$$

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  • $\begingroup$ @snulty Thank you very much. That's a serious typo and I will edit now. $\endgroup$ – DonAntonio Mar 31 '16 at 18:10
  • $\begingroup$ you're welcome :) $\endgroup$ – snulty Mar 31 '16 at 18:13

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