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

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    @Vikram: It was just a minor slip, you were concentrating on the top. But actually it was the bottom that mattered all along.2012-04-07

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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.