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Let $a_n=1+\frac{1}{2^2}+\dots+\frac{1}{n^2}$. Considering known that $\lim_{n\to\infty}a_n=\frac{\pi^2}{6}$, evaluate $$\lim_{n\to\infty}n\left( a_n-\frac{\pi^2}{6}\right)$$ My attempt: I first proved that $b_n=n\left( a_n-\frac{\pi^2}{6}\right)$ is decreasing , i.e. $b_{n+1}-b_n=a_n+\frac{1}{n+1}-\frac{\pi^2}{6} \leq 0$, which is true since $$\lim_{n\to\infty}\left(a_n+\frac{1}{n+1}\right)=\frac{\pi^2}{6}$$ and $a_n+\frac{1}{n+1}$ is an increasing sequence.

Then I proved that $b_n$ is bounded by $0$ (obviously) and $-1$, the latter being true since it is equivalent to $a_n+\frac{1}{n}\geq \frac{\pi^2}{6}$, which can be proven as above.

So $b_n$ is both decreasing and bounded, which means that it has a limit $l$ and I ended with Stolz-Cesaro: $$l=\lim_{n\to\infty}n\left( a_n-\frac{\pi^2}{6}\right)=\lim_{n\to\infty}\frac{n^2\left( a_n-\frac{\pi^2}{6}\right)}{n}=\lim_{n\to\infty}\left( (n+1)^2\left( a_{n+1}-\frac{\pi^2}{6}\right) -n^2\left( a_n-\frac{\pi^2}{6}\right)\right)=\lim_{n\to\infty}\left( \frac{n^2}{(n+1)^2}+a_{n+1}-\frac{\pi^2}{6}+2n\left( a_{n+1}-\frac{\pi^2}{6}\right)\right)=1+0+2l$$ so $l=-1$.

Can anyone provide a shorter solution, if there is one, please?

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Note that

$$\frac{\pi^2}6-a_n=\sum_{k=n+1}^\infty\frac1{k^2}$$

And

$$\frac1{n+1}=\int_{n+1}^\infty\frac1{x^2}~dx<\sum_{k=n+1}^\infty\frac1{k^2}<\int_n^\infty\frac1{x^2}~dx=\frac1n$$

Therefore, by the squeeze theorem, the limit is $-1$.


If one wishes to avoid integrals, creative telescoping may be employed:

$$\frac1k-\frac1{k+1}=\frac1{k(k+1)}<\frac1{k^2}<\frac1{(k-1)k}=\frac1{k-1}-\frac1k$$

I leave the telescoping process to the reader. $\ddot\smile$

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You may like this. By the Stolz-Cesaro Theorem, one has \begin{eqnarray} &&\lim_{n\to\infty}n\left( a_n-\frac{\pi^2}{6}\right)\\ &=&\lim_{n\to\infty}\frac{ a_n-\frac{\pi^2}{6}}{\frac1n}\\ &=&\lim_{n\to\infty}\frac{(a_{n+1}-\frac{\pi^2}{6})- (a_n-\frac{\pi^2}{6})}{\frac1{n+1}-\frac1n}\\ &=&\lim_{n\to\infty}\frac{a_{n+1}-a_n}{-\frac1{n(n+1)}}\\ &=&\lim_{n\to\infty}\frac{\frac1{(n+1)^2}}{-\frac1{n(n+1)}}\\ &=&-1. \end{eqnarray}

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  • $\begingroup$ I do like this (+1) $\endgroup$ – Simply Beautiful Art Sep 6 '17 at 20:21
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    $\begingroup$ But in order to apply Stolz-Cesaro, don't we need the sequence on the denomitor to be monotonic and unbounded? $\frac{1}{n}$ is bounded. $\endgroup$ – Shroud Sep 6 '17 at 20:26
  • $\begingroup$ D: Good catch @Shroud $\endgroup$ – Simply Beautiful Art Sep 6 '17 at 20:29
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    $\begingroup$ I haven't used Stolz-Cesàro many times, so I've learned something new today! =D $\endgroup$ – Simply Beautiful Art Sep 6 '17 at 20:36
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    $\begingroup$ @SimplyBeautifulArt : the proof for $0/0$ case is not in wiki but luckily MSE has it: math.stackexchange.com/a/631160/72031 $\endgroup$ – Paramanand Singh Sep 6 '17 at 22:52

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