I have an exam on monday and I just want to make sure I am using this notation correctly. Is there any ways I can make this argument style more rigorous?
It seems a bit hand wavey using this notation instead of directly applying Taylor's theorem in each case. For example, I feel ok writing: $\sin(x)=x-\frac{x^3}{3!}+\frac{\cos^5(\theta)}{5!}x^5$ for some $\theta\in(0,x)$.
When I am writing $\sin(x)=x-\frac{x^3}{3!}+O(x^5)$ I am just copying other people who are successful and am not entirely sure what theorem I am quoting.
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Here is how I did it for the first series, $\sum (n^{1/n}-1)$
We examine the terms $a_n=(n^{1/n}-1) =e^{\frac{1}{n}\log(n)}-1=(1+\frac{1}{n}\log(n)+O(\frac{1}{n^2}\log(n)^2))-1=\frac{1}{n}\log(n) +O(\frac{1}{n^2}\log(n)^2))\geq \frac{1}{n}$
for all positive $n$ which make the expression defined. Therefore it diverges by the comparison test.
The next series, $\sum \log(1+\frac{1}{n\sqrt{n}})$:
Examining the terms $a_n=\log(1+\frac{1}{n\sqrt{n}})=(\frac{1}{n^{3/2}}+O(\frac{1}{n^3}))$
Now using the limit comparison test with $b_n=\frac{1}{n^{3/2}}$, we take the limit $\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty} \frac{\frac{1}{n^{3/2}}+O(\frac{1}{n^3})}{\frac{1}{n^{3/2}}}=\lim_{n\to\infty}=\frac{1+O(\frac{1}{n^{3/2}})}{1}=1<+\infty$
Therefore, $\sum \log(1+\frac{1}{n\sqrt{n}})$ converges by the limit comparison test.
I hope I at least have the correct conclusions for these series...
Any feedback is very welcomed and thanks in advance!