# How to go from $\lim_{n\to \infty}\frac{81}4(\frac{n(n+1)}{2})^2-\frac{54n(n+1)}{2n^2}$ to $\lim_{n\to \infty} \frac{81}4(\frac1n+1)^2-27(\frac1n+1)$?

It was a question in a calculus textbook and I didn't know how he did the algebra between these steps

$$\lim_{n\to \infty} \frac{81}{4}\left(\frac{n(n+1)}{2}\right)^2-\frac{54}{n^2}\frac{n(n+1)}{2}$$

$$\lim_{n\to \infty} \frac{81}{4}\left(\frac{1}{n}+1\right)^2-27\left(\frac{1}{n}+1\right)$$

How did he go from the first to the second step?

• You mean $n\to\infty$, right? – draks ... Nov 22 '13 at 11:39
• Limit should be on $n$ first part is wrong. – lab bhattacharjee Nov 22 '13 at 11:39
• $\dfrac{54}{2}=27$; $\dfrac{n(n+1)}{n^2}=\dfrac{1}{n}+1$. But, as written, the two expressions are not equal. – egreg Nov 22 '13 at 11:42

HINT: $\hskip1.7in\displaystyle\frac{n(1+n)}{n^2}=\frac{n+n^2}{n^2}=\frac n{n^2} + \frac{n^2}{n^2}=\frac1n+1,$
so it looks like there's a $n^4$ ($n^2$ inside the square's bracket) missing in the first denominator...