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is the sequence

$${ a }_{ n }=\frac { { \left( -1 \right) }^{ n+1 }n }{ n+\sqrt { n } } $$ convergent or divergent?

I used comparison test, and I compared it with $\frac{n}{n+\sqrt{n}}$. As $\lim _{ n\rightarrow \infty }{ \frac { n }{ n+\sqrt { n } } =1 } $. So ${ a }_{ n }=\frac { { \left( -1 \right) }^{ n+1 }n }{ n+\sqrt { n } } $ is convergent. Is this correct?

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    $\begingroup$ @BobHanlon Sequence $\neq$ series $\endgroup$
    – Clement C.
    Jul 28, 2016 at 21:28
  • $\begingroup$ The 'comparison test' does not say what you thought/guessed/hoped it says. $\endgroup$
    – user21820
    Jul 31, 2016 at 6:42

3 Answers 3

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This is not correct. One may observe that $$ \left.\frac{(-1)^{n+1}n}{n+\sqrt{n}}\right|_{n=2p}=\frac{-2p}{2p+\sqrt{2p}} \to -1 \quad \text{as}\quad p \to \infty $$and $$ \left.\frac{(-1)^{n+1}n}{n+\sqrt{n}}\right|_{n=2p+1}=\frac{2p+1}{2p+1+\sqrt{2p+1}} \to 1 \quad \text{as}\quad p \to \infty $$

The given sequence is not convergent.

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No the sequence does not converge. Consider, along even numbers $n = 2k$, we have $$\frac{(-1)^{2k+1}(2k)}{2k + \sqrt{2k}} = - \frac{1}{1 + \frac{1}{\sqrt{2k}}} \to -1 \,\,\, \text{ as } k \to \infty$$ where the first equality comes from multiplying the top and bottom by $2k$. Whereas, along odd numbers $n = 2m + 1$, we have $$\frac{(-1)^{2m+2}(2m+1)}{2m+1 + \sqrt{2m+1}} = \frac{1}{1 + \frac{1}{\sqrt{2m+1}}} \to 1 \,\,\, \text{ as } m \to \infty$$ where again, the first inequality comes from multiplying the top and bottom by $2m+1$. Since there are two subsequences which converge to different limits, the sequence does not converge.

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The absolute values of the terms are $ \dfrac n {n+\sqrt n} $, and those approach $1$ as $n\to\infty$. The factor $(-1))^n$ alternates between $1$ and $-1$, so you have the even-numbered terms approaching $1$ and the odd-numbered terms approaching $-1$. Consequently the whole sequence does not approach a particular number.

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