Argue that the sequence converges 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.
Hint: $a_n=n^2\cos(1/n)-n^2 = -n^2 2\sin^2 \frac{1}{2n}\to -\frac{1}{2}, n \to \infty$
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&amp;=n^2(\cos\frac1n-1)=-n^2\,\frac{\sin^2\frac1n}{\cos\frac1n+1}\\[0.3cm]
&amp;=-\frac1{\cos\frac1n+1}\,\Bigg(\frac{\sin\frac1n}{\frac1n}\Bigg)^2\xrightarrow[n\to\infty]{}-\frac12\times1=-\frac12.
\end{align}