Resolving a list of exercise these two exercises I could not
Determined using the comparison test
1) $\displaystyle \sum_{n=1}^{\infty} \frac{(n+1)}{n^3}$
2) $\displaystyle \sum_{n=0}^{\infty} \frac{n^2+1}{n^3+1}$
The limit comparison test says that:
- If the limit of a[n]/b[n] is positive, then the sum of a[n] converges if and only if the sum of b[n] converges.
- If the limit of a[n]/b[n] is zero, and the sum of b[n] converges, then the sum of a[n] also converges.
- If the limit of a[n]/b[n] is infinite, and the sum of b[n] diverges, then the sum of a[n] also diverges.
For the first one, compare it to $\displaystyle \frac{1}{n^2}$. $\displaystyle \lim_{n\to\infty}\frac{\frac{(n+1)}{n^3}}{\frac{1} {n^2}} = 1$, so the sum converges because $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ does.
For the second one, compare it to $\displaystyle \frac{1}{n+1}$. $\displaystyle \lim_{n\to\infty}\frac{\frac{n^2+1}{n^3+1}}{\frac{ 1}{n+1}}=1$, so the sum diverges because $\displaystyle \sum_{n=0}^{\infty}\frac{1}{n+1}$ does.
i read comparison test, so we can actually use the direct comparison test.
1) $\displaystyle \frac{n+1}{n^{3}}\le \frac{2n}{n^{3}}=\frac{2}{n^{2}}.$
2) $\displaystyle \frac{n^{2}+1}{n^{3}+1}>\frac{n^{2}}{2n^{3}}=\frac {1}{2n}.$
As long as $\displaystyle n\ge1.$
I have a doubt.
If I have one series $\displaystyle a_n$ I want to determine the convergence, found that the series $\displaystyle b_n$ is convergent, but $\displaystyle b_n$ > $\displaystyle a_n$ up to a value of n, not for all values of n. The series $\displaystyle a_n$ is convergent ?
Not necessarily. If you are comparing $\displaystyle a_n$ to $\displaystyle b_n$ and you know that $\displaystyle b_n$ converges, $\displaystyle a_n\leq b_n$ for all but finitely many $\displaystyle n$ will tell you that $\displaystyle a_n$ converges.
For example, say $\displaystyle b_n$ converges. You compare $\displaystyle a_n$ to $\displaystyle b_n$ and find:
1.) $\displaystyle \left\{\begin{array}{lr}a_n<b_n &: n\leq 1000\\a_n>b_n &: n>1000\end{array}\right\}$ $\displaystyle a_n$ does not necessarily converge.
2.) $\displaystyle \left\{\begin{array}{lr}a_n>b_n &: n\leq 1000\\a_n<b_n &: n>1000\end{array}\right\}$ $\displaystyle a_n$ will converge.
Note that my choice of 1000 was arbitrary.
$\displaystyle \lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = \lim_{n\to\infty}\frac{n}{\sqrt{n^2-1}} = \lim_{n\to\infty}\frac{n}{n\sqrt{1-1/n^2}} = \lim_{n\to\infty}\frac{1}{\sqrt{1-1/n^2}}=\frac{1}{\sqrt{1-0}}=1$
So $\displaystyle a_n$ converges because $\displaystyle b_n$ does.