Argue that the sequence converges

Question

Let $$a_n=n^2\cos(1/n)-n^2$$ Show that the sequence converges.

Now, I know how to use the formal definition of convergence but I am looking for simpler methods (i.e the tests for series). I found it to converge to $-1/2$ but I need some method to show that it actually converges. Same problem applies to the sequence (not series) $$\frac{n^3}{n^2}-\frac{n^3}{(n+1)^2} $$

I know the process of finding the value as $n$ shoots off to infinity for this sequence would just be write it all under 1 fraction and divide by the biggest power of $n$. But I run into the same problem of how to argue that it does indeed converge.

Answer 1

Hint: $a_n=n^2\cos(1/n)-n^2 = -n^2 2\sin^2 \frac{1}{2n}\to -\frac{1}{2}, n \to \infty$

Answer 2

They are not similar problems. For $$ \frac{n^3}{n^2}-\frac{n^3}{(n+1)^2}, $$ you just do a tiny bit of elementary manipulation to write it as $$ \frac{n(n+1)^2-n^3}{(n+1)^2}=\frac{2n^2+n}{(n+1)^2}=\frac{2+1/n}{(1+1/n)^2}\xrightarrow[n\to\infty]{}2 $$

For the cosine, you need to use some property specific to the cosine. If you know that $\cos(t)=1-\frac{t^2}2+o(t^4)$, you get $$ n^2\cos\frac1n-n^2=\frac12+o(\frac1{n^2})\xrightarrow[n\to\infty]{}0. $$ If for instance you know that $\lim_{t\to0}\frac{\sin t}t=1$, you can do, from $\cos^2t-1=(\cos t -1)(\cos t+1)$, $$ \cos t -1 =\frac{\cos^2t-1}{\cos t+1}=-\frac{\sin^2t}{\cos t+1}. $$ Then \begin{align} n^2\cos\frac1n-n^2&=n^2(\cos\frac1n-1)=-n^2\,\frac{\sin^2\frac1n}{\cos\frac1n+1}\\[0.3cm] &=-\frac1{\cos\frac1n+1}\,\Bigg(\frac{\sin\frac1n}{\frac1n}\Bigg)^2\xrightarrow[n\to\infty]{}-\frac12\times1=-\frac12. \end{align}