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a nice problem #2

sbhatnagar

Active member
Jan 27, 2012
95
Find the limit.

$$\lim_{n \to \infty} \left\{ \left( 1+\frac{1^2}{n^2}\right)\left( 1+\frac{2^2}{n^2}\right)\left( 1+\frac{3^2}{n^2}\right) \cdots \left( 1+\frac{n^2}{n^2}\right)\right\}^{\dfrac{1}{n}}$$
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
If $L$ is the limit, then $\log L=\displaystyle\lim_{n \to{+}\infty}\dfrac{1}{n}\;\displaystyle\sum_{k=1}^n \log \left(1+(k/n)^2\right)=\displaystyle\int_0^1 \log (1+x^2)\;dx$ .
 

sbhatnagar

Active member
Jan 27, 2012
95
If $L$ is the limit, then $\log L=\displaystyle\lim_{n \to{+}\infty}\dfrac{1}{n}\;\displaystyle\sum_{k=1}^n \log \left(1+(k/n)^2\right)=\displaystyle\int_0^1 \log (1+x^2)\;dx$ .
Yeah that's right. The integral can be further simplified. \( \displaystyle \int_{0}^{1}\ln(1+x^2) dx= \cdots=\frac{\pi}{2}-2+\ln(2)\).

Therefore \( \displaystyle L=2+e^{\dfrac{\pi}{2}-2}\).
 

Sherlock

Member
Jan 28, 2012
59
The integral is just boring old integration by parts (the real key part was realizing that it's a Riemann sum). I'm sure that's why Dr Revilla didn't bother to provide us with the calculation. But we can make it interesting in such a way that the given problem is in the end done by writing it as sum then as an integral, then as a sum, and then finally as an integral. You don't believe me? Watch this:

$$\displaystyle \begin{aligned} & \begin{aligned} \log (\ell) & = \lim_{n \to{+}\infty}\dfrac{1}{n}\;\displaystyle\sum_{k=1} ^n \log \left(1+(k/n)^2\right) \\& = \int_0^1 \log (1+x^2)\;dx \\& = \int_0^1\sum_{k \ge 0}\frac{(-1)^kx^{2k+2}}{k+1}\;{dx} \\& = \sum_{k \ge 0}\int_{0}^{1}\frac{(-1)^kx^{2k+2}}{k+1}\;{dx} \\& = \sum_{k \ge 0}\bigg[\frac{(-1)^kx^{2k+3}}{(k+1)(2k+3)}\bigg]_0^1 \\& = \sum_{k \ge 0}\frac{(-1)^k}{(k+1)(2k+3)} \\& = \sum_{k \ge 0}\frac{(-1)^k(2k+3)-2(-1)^k(k+1)}{(k+1)(2k+3)} \\& = \sum_{k\ge 0}\bigg(\frac{(-1)^k}{k+1}-\frac{2(-1)^k}{2k+3}\bigg) \\& = \sum_{k\ge 0}\int_0^1\bigg((-1)^kx^{k}-2(-1)^kx^{2k+2}\bigg)\;{dx} \\& = \int_0^1\sum_{k\ge 0}\bigg( (-1)^kx^{k}-2(-1)^kx^{2k+2}\bigg)\;{dx} \\&= \int_0^1\bigg(\frac{1}{1+x}-\frac{2x^2}{1+x^2}\bigg)\;{dx} \\& = \int_0^1\bigg(\frac{1}{1+x}-\frac{2x^2+2-2}{1+x^2}\bigg)\;{dx} \\& = \int_0^1\bigg(\frac{1}{1+x}-2+\frac{2}{1+x^2}\bigg)\;{dx} \\& = \ln|1+x|-2+2\tan^{-1}{x}\bigg|_{0}^{1} \\& = \ln(2)-2+\frac{\pi}{2} \end{aligned} \\& \therefore ~ \ell = \exp\left(\ln{2}-2+\frac{\pi}{2}\right). \end{aligned}$$
 

Krizalid

Active member
Feb 9, 2012
118
I like the double integration approach:

$\begin{aligned} \int_{0}^{1}{\ln (1+{{x}^{2}})\,dx}&=\int_{0}^{1}{\int_{0}^{{{x}^{2}}}{\frac{dt\,dx}{1+t}}} \\
& =\int_{0}^{1}{\int_{\sqrt{t}}^{1}{\frac{dx\,dt}{1+t}}} \\
& =\int_{0}^{1}{\frac{1-\sqrt{t}}{1+t}\,dt} \\
& =2\int_{0}^{1}{\frac{t-{{t}^{2}}}{1+{{t}^{2}}}\,dt} \\
& =-2+\frac{\pi }{2}+\ln 2.
\end{aligned}
$

Now, integration by parts it's faster than any of these methods, perhaps you may consider it a boring way, but it's still valid and saves a lot of time.