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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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    If you got $e^{-1}$ as $n\to\infty$ you did something wrong.2012-04-07
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    Not that it matters, but it should be $n^n$ in the denominator, not $n^{2n}$.2012-04-07
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    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.2012-04-07
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    @Andre, can you pls throw some light on why the denominator has to be n^n, any article.2012-04-07
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    No article needed here, you simply made the mistake of replacing $(n-1)/n^2$ by $(1-1/n)/n^2$ instead of $(1-1/n)/n$.2012-04-07
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    @Didier , thanx2012-04-07
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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.