# Root test for series

Find if the series converges or diverges, $a_n=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n^2}\right)^n$, simplifying the series expression we get $\left(\frac{n-1}{n^2}\right)^n=\frac{(1+(-1)/n)^n}{(n)^{2n}}$, conducting Root test, taking nth root of the simplified expression as $n\to\infty$, $e^{-1}$ is this methods correct? OR as the author has done by taking the nth root of the original expression of $a_n$, we get $\lim_{n\to\infty}(\frac{1}{n}-\frac{1}{n^2})=0 \Rightarrow a_n$ converges.

-
If you got $e^{-1}$ as $n\to\infty$ you did something wrong. –  anon Apr 7 '12 at 6:21
Not that it matters, but it should be $n^n$ in the denominator, not $n^{2n}$. –  André Nicolas Apr 7 '12 at 7:17
Your $e^{-1}$ is the limit of the $n$th root of the numerator only, you should be computing the $n$th root of the fraction. The denominator should be $n^n$ (and not your $n^{2n}$) and its $n$th root is $n$, hence the $n$th root of the fraction is $\sim1/(en)$, which goes to zero. –  Did Apr 7 '12 at 8:23
@Andre, can you pls throw some light on why the denominator has to be n^n, any article. –  Vikram Apr 7 '12 at 11:37
No article needed here, you simply made the mistake of replacing $(n-1)/n^2$ by $(1-1/n)/n^2$ instead of $(1-1/n)/n$. –  Did Apr 7 '12 at 11:42

You are right that $\lim\limits_{n\to\infty} \left(1-\frac1n\right)^n=e^{-1}$. But this is not the limit you use in the root test. You are looking for the value of $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$. (You worked only with the denominator of your expression for $a_n$, but you have to work with the whole expression and take the $n$-th root.)
The limits $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$ is exactly $\lim\limits_{n\to\infty} \left(\frac1n-\frac1{n^2}\right)=0$.
You will get the same value from $$\sqrt[n]{a_n}=\sqrt[n]{\frac{(1+(-1)/n)^n}{n^{n}}} = \frac{1-\frac1n}{n}$$ which tends to $0$ as $n\to\infty$.
This implies that the series $\sum a_n$ converges.