leading terms in sequence What is the leading term of the following sequences as $n \to \infty$

(a) $n^{-1}- (n^{2}+1)^{-1/2}$
(a) $7^{1/n^2}- 2(7^{2/n^2})+7^{3/n^2}$


 A: We  expand  the expression in  a  Taylor series around $n=0$ and extract the most significant term when $n\rightarrow \infty$

We obtain
  \begin{align*}
\color{blue}{n^{-1}-\left(n^2+1\right)^{-\frac{1}{2}}}
&=\frac{1}{n}-\frac{1}{n}\left(1+\frac{1}{n^2}\right)^{-\frac{1}{2}}\\
&=\frac{1}{n}-\frac{1}{n}\sum_{j=0}^\infty\binom{-\frac{1}{2}}{j}\frac{1}{n^{2j}}\tag{1}\\
&=\frac{1}{n}-\frac{1}{n}\left(1+\left(-\frac{1}{2}\right)\frac{1}{n^2}
+\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\frac{1}{n^4}+\cdots\right)\\
&=\frac{1}{2n^3}-\frac{3}{4n^5}-\cdots\\
&\color{blue}{\sim \frac{1}{2n^3}}
\end{align*}

Comment:


*

*In (1) we use the binomial series expansion.



We obtain
  \begin{align*}
\color{blue}{7^{1/n^2}}&\color{blue}{-2\cdot7^{2/n^2}+7^{3/n^2}}\\
&=e^{\frac{1}{n^2}\ln 7}-2e^{\frac{2}{n^2}\ln 7}+e^{\frac{3}{n^2}\ln 7}\tag{2}\\
&=\left(1+\frac{\ln 7}{n^2}+\frac{1}{2}\cdot\frac{\ln^2 7}{n^4}+\cdots\right)
-2\left(1+\frac{2\ln 7}{n^2}+\frac{1}{2}\cdot\frac{4\ln^2 7}{n^4}+\cdots\right)\tag{3}\\
&\qquad +\left(1+\frac{3\ln 7}{n^2}+\frac{1}{2}\cdot\frac{9\ln^2 7}{n^4}+\cdots\right)\\
&=\left(\frac{1}{2}-4+\frac{9}{2}\right)\frac{\ln^2 7}{n^4}+\cdots\\
&\color{blue}{\sim \frac{\ln^2 7}{n^4}}
\end{align*}

Comment:


*

*In (2) we use the representation $a^b=e^{b\ln(a)}$.

*In (3) we use the exponential series expansion $$e^x=\sum_{j=0}^\infty\frac{x^j}{j!}=1+x+\frac{1}{2}x^2+\cdots$$

Author's example:
Let's verify the author's calculation. We recall the cosine series expansion at $x=0$ is
  \begin{align*}
\sum_{j=0}^\infty(-1)^j\frac{x^{2j}}{(2j)!}
\end{align*}
  It is also convenient to use the  Landau Big-O notation.
We  obtain
  \begin{align*}
\color{blue}{1-\cos\left(\frac{2n^{1/2}+3}{3n^2+1}\right)}
&=1-\cos\left(\frac{\frac{2}{3}n^{-3/2}+n^{-2}}{1+\frac{1}{3}n^{-2}}\right)\tag{5}\\
&=1-\cos\left(\left(\frac{2}{3}n^{-3/2}+O(n^{-2})\right)\left(1+O(n^{-2})\right)\right)\tag{6}\\
&=1-\cos\left(\frac{2}{3}n^{-3/2}+O(n^{-2})\right)\tag{7}\\
&=1-\left(1-\frac{1}{2}\left(\frac{2}{3}n^{-3/2}+O(n^{-2})\right)^2+O(n^{-6})\right)\\
&=\frac{1}{2}\cdot\frac{4}{9}n^{-3}+O(n^{-7/2})\\
&\color{blue}{\sim \frac{2}{9}n^{-3}}
\end{align*}
  confirming the author's result.

Comment:


*

*In (5) we factor out $3n^2$ as preparation for a geometric series expansion.

*In (6) we do the series expansion noting that
$$\frac{1}{1+\frac{1}{3}n^{-2}}=1-\frac{1}{3}n^{-2}+\frac{1}{9}n^{-4}-\cdots=1+O(n^{-2})$$

*In (7) we observe that terms are swallowed by $O(n^{-2})$.
