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These limits give me a real hard time. I tried limit comparison and squeeze theorems to no avail.

$$\lim_{n \to \infty} \frac{\sum^{n}_{k=1} k^2\tan^{-1}k}{\sum^n_{k=0}(n+k)^2} = \space ?$$

$$\lim_{n \to \infty} \frac{1}{n} \sum^{n}_{k=1} \ln\left(\frac{2k-1}{2k}\right) = \space?$$

What are some other techniques i might use to solve these?

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  • $\begingroup$ For the first, we can use that fact that pretty quickly $\arctan k$ is fairly close to $\pi/2$. Then use Squeezing. $\endgroup$ – André Nicolas Mar 10 '14 at 21:54
  • $\begingroup$ See Riemann sum. $\endgroup$ – Lucian Mar 10 '14 at 21:59
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$$\lim_{n\to\infty}\frac{\displaystyle\sum_{k=0}^nk^2\arctan k}{\displaystyle\sum_{k=0}^n(n+k)^2}=\lim_{n\to\infty}\frac{\displaystyle\frac1{n^3}\cdot\sum_{k=0}^nk^2\arctan k}{\displaystyle\frac1{n^3}\cdot\sum_{k=0}^n(n+k)^2}=\lim_{n\to\infty}\frac{\displaystyle\frac1n\cdot\sum_{k=0}^n\bigg(\frac kn\bigg)^2\arctan k}{\displaystyle\frac1n\cdot\sum_{k=0}^n\bigg(1+\frac kn\bigg)^2}=$$

$$=\frac{\displaystyle\lim_{n\to\infty}\frac1n\cdot\sum_{k=0}^n\bigg(\frac kn\bigg)^2\arctan k}{\displaystyle\lim_{n\to\infty}\frac1n\cdot\sum_{k=0}^n\bigg(1+\frac kn\bigg)^2}{\large\approx}\frac{\displaystyle\frac\pi2\cdot\int_0^1x^2\,dx}{\displaystyle\int_0^1(1+x)^2\,dx}=\frac{\dfrac\pi2\cdot\bigg[\dfrac{x^3}3\bigg]_0^1}{\bigg[\dfrac{(1+x)^3}3\bigg]_0^1}=\frac\pi{14}$$


For the second sum, use the fact that $\ln(1+x)$ $\large\approx$ $x$ when $x\to0$. In our case, $x=-\dfrac1{2k}$ . Then use the fact that $\displaystyle\sum_{k=1}^n\frac1k{\large\approx}\ln n$. And finally, $\displaystyle\lim_{n\to\infty}\frac{\ln n}n=0$.

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