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Set the sequence $a_n$ such that $\{a_n\}=\dfrac{(-1)^{n-1}n}{n^2+2}$. If $|a_n|$ converges (only to $0$, it would seem; correct me if I'm wrong), then $a_n$ must too converge, both to some value $L = \lim a_n = \lim |a_n|$.

So, $\displaystyle \lim a_n = \lim (-1)^{n-1}\frac{n}{n^2+2} = \lim |(-1)^{n-1}\frac{n}{n^2+2}| = \lim \frac{n}{n^2+2}$.

By the theorem which says a series $a_n = f(n)$ must converge to the same value of its corresponding real-valued function ($f(x)$), it may be concluded that if $\displaystyle \lim_{x \to \infty}\frac{x}{x^2+2}$ converges to some value $L$, then $\displaystyle \lim_{x \to \infty}\frac{x}{x^2+2} = L \implies \lim\frac{n}{n^2+2} = L$, and thus $a_n$ converges to $L$.

So, $\displaystyle \lim_{x \to \infty}\frac{x}{x^2+2}$ results in the indeterminate form $\dfrac{\infty}{\infty}$, and this allows the use of L'Hôpital's rule to find the limit $\displaystyle \lim_{x \to \infty}\frac{x}{x^2+2}$ to equal $\displaystyle \lim_{x \to \infty}\frac{1}{2x}=0$.

Therefore, $a_n$ converges to $0$.

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  • $\begingroup$ Do you have a question? $\endgroup$
    – abiessu
    Jul 8, 2015 at 1:30
  • $\begingroup$ @abiessu See tag: solution-verification $\endgroup$
    – alxmke
    Jul 8, 2015 at 1:32
  • $\begingroup$ @abiessu Does noting that clear things up? $\endgroup$
    – alxmke
    Jul 8, 2015 at 1:38
  • $\begingroup$ "then $a_n$ must too converge, both to some value $L = \lim a_n = \lim |a_n|$." I don't see how this follows without more assumptions. Consider the constant sequence $a_n = -1$ for all $n.$ Then $|a_n| = 1$ converges to $1$ but $a_n$ converges to $-1$. $\endgroup$
    – R R
    Jul 8, 2015 at 1:53
  • $\begingroup$ @RR That's why I said "only to 0, it would seem; correct me if I'm wrong", because I remember something along those lines. But yes, I too had considered that case. $\endgroup$
    – alxmke
    Jul 8, 2015 at 1:57

1 Answer 1

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What can really be said except that it is correct. Although, it could have been achieved more directly by noting: $$ \lim_{n \to \infty} \frac{n (-1)^{n-1}}{n^2 + 2} = \lim_{n \to \infty} \frac{ (-1)^{n-1} / n}{1 + 2/n^2} = \frac{ \lim_{n \to \infty} \left\{ (-1)^{n-1} / n \right\}}{1 + \lim_{n \to \infty} \{ 2/n^2 \} } = 0 . $$

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