# Determining whether the given series is convergent or divergent.

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

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 $(1)$.

• And how did you get that expression from the final sum? That is not obvious. – Rory Daulton Mar 31 '16 at 17:57
• @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 $x=1/6$. – Mark Viola Mar 31 '16 at 18:00

The $\;n$ - th root test as this is a positive series

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

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