Master limit 2

Calculus Level 4

$\displaystyle\lim \limits_{n\to \infty }\displaystyle\sum \limits^{n}_{k=1}\frac{\sqrt{\dbinom{n + k}{ 2} } }{n^{2}} = \ ?$

\frac {3\sqrt {2}}{4}

\frac {1}{2}

\frac {\sqrt {2}}{4}

\infty

2 solutions

Kenny Lau
Oct 12, 2015

This contains one fallacious step... yet...

$\displaystyle\quad\lim_{n\to\infty}\sum_{k=1}^n\frac{\sqrt{\binom{n+k}{2}}}{n^2}$

$=\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac{\sqrt{\frac{(n+k)(n+k-1)}{2}}}{n^2}$

$=\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac1n\sqrt{\frac{(1+\frac kn)(1+\frac kn-\frac1n)}{2}}$

$=\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac1n\sqrt{\frac{(1+\frac kn)(1+\frac kn)}{2}}$

$=\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac1n\frac{1+\frac kn}{\sqrt2}$

$=\displaystyle\int_0^1\frac{1+x}{\sqrt2}dx$

$=\dfrac{3\sqrt2}4$

Yep... You need to solve this by Squeeze Theorem.

Pi Han Goh - 5 years, 6 months ago

Carlos Victor
Jun 12, 2016

I used the Theorem Stolz-Cesaro.

Can you upload your solution

rishabh singhal - 4 years, 10 months ago